blocks|key|2065707|text|Python+2，24字节|type|header-one|depth|inlineStyleRanges|entityRanges|offset|length|data|2065708|lambda+n,k:n*10%25k==0|code-block|syntax|javascript|2065709|在网上试试！|unstyled|style|CODE|2065710|检查10n是k的倍数，但n本身不是。|2065711|24字节|BOLD|2065712|lambda+n,k:n%25k>>n*10%25k*n|2065713|2065714|不能省略结尾处的*n，比如n=19001,+k=10000。|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/2/|1|https://tio.run/##K6gsycjPM/qfZhvzPycxNyklUSFPJ9sqT8vQQDXb1tbAJk81@395RmZOqoKhFRcnUE7BViEzr6C0REOTi7OgKDOvRCFNAyis@T/aQkfBJJYLRJkCKUsIZa6jYBwLAA|2|https://tio.run/##K6gsycjPM/qfZhvzPycxNyklUSFPJ9sqTzXbzi5Py9BANVsr7395RmZOqoKhFRcnUE7BViEzr6C0REOTi7OgKDOvRCFNAyis@T/aQkfBJJYLRJkCKUsIZa6jYAykDC0NDAx1FAwNgCAWAA^0|0|8|0|0|0|0|6|0|6|1|0|0|I|0|0|4|0|4|0|0|0|6|0|6|2|0|0|U^^$0|@$1|2|3|4|5|6|7|18|8|@]|9|@$A|19|B|1A|1|1B]]|C|$]]|$1|D|3|E|5|F|7|1C|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|1D|8|@$A|1E|B|1F|L|M]]|9|@$A|1G|B|1H|1|1I]]|C|$]]|$1|N|3|O|5|K|7|1J|8|@$A|1K|B|1L|L|M]]|9|@]|C|$]]|$1|P|3|Q|5|K|7|1M|8|@$A|1N|B|1O|L|M]|$A|1P|B|1Q|L|R]]|9|@]|C|$]]|$1|S|3|T|5|F|7|1R|8|@]|9|@]|C|$G|H]]|$1|U|3|J|5|K|7|1S|8|@$A|1T|B|1U|L|M]]|9|@$A|1V|B|1W|1|1X]]|C|$]]|$1|V|3|W|5|K|7|1Y|8|@$A|1Z|B|20|L|M]]|9|@]|C|$]]]|X|$Y|$5|Z|10|11|C|$12|13]]|14|$5|Z|10|11|C|$12|15]]|16|$5|Z|10|11|C|$12|17]]]]

<h1><a href="https://docs.python.org/2/" rel="noreferrer">Python 2</a>, 24 bytes</h1>

<pre class="lang-python prettyprint-override"><code>lambda n,k:n*10%k==0&lt;n%k
</code></pre>
<p><a href="https://tio.run/##K6gsycjPM/qfZhvzPycxNyklUSFPJ9sqT8vQQDXb1tbAJk81@395RmZOqoKhFRcnUE7BViEzr6C0REOTi7OgKDOvRCFNAyis@T/aQkfBJJYLRJkCKUsIZa6jYBwLAA" rel="noreferrer" title="Python 2 – Try It Online">Try it online!</a></p>
<p>Checks that <span class="math-container">\$10n\$</span> is a multiple of <span class="math-container">\$k\$</span>, but <span class="math-container">\$n\$</span> itself is not.</p>
<p><strong>24 bytes</strong></p>

<pre class="lang-python prettyprint-override"><code>lambda n,k:n%k&gt;&gt;n*10%k*n
</code></pre>
<p><a href="https://tio.run/##K6gsycjPM/qfZhvzPycxNyklUSFPJ9sqTzXbzi5Py9BANVsr7395RmZOqoKhFRcnUE7BViEzr6C0REOTi7OgKDOvRCFNAyis@T/aQkfBJJYLRJkCKUsIZa6jYAykDC0NDAx1FAwNgCAWAA" rel="noreferrer" title="Python 2 – Try It Online">Try it online!</a></p>
<p>The <code>*n</code> at the end can't be omitted, say for <code>n=19001, k=10000</code>.</p>


blocks|key|2065823|text|R，31字节|type|header-one|depth|inlineStyleRanges|entityRanges|offset|length|data|2065824|function(x,y)nchar((x/y)%25%251)==3|code-block|syntax|javascript|2065825|在网上试试！|unstyled|2065826|小数部分(%25%251)的x/y必须是三个字符:+so，0.1，0.2+.+0.9，而不是0或0.3333。|style|CODE|entityMap|0|LINK|mutability|MUTABLE|url|https://www.r-project.org/|1|https://tio.run/##K/qfoJpZHF@cmpuZklmWWZyZlJMan1SpmmD7P600L7kkMz9Po0KnUjMvOSOxSEOjQr9SU1XVUNPW1vi/mQJWnQomCsoKJUWlqVwWeBSkJeYU41ZhCjPCkpACcxwKjOF2/AcA^0|0|1|0|0|0|0|6|1|0|5|3|A|3|P|3|T|3|Z|3|16|1|18|6^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@$A|Z|B|10|1|11]]|C|$]]|$1|D|3|E|5|F|7|12|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|13|8|@]|9|@$A|14|B|15|1|16]]|C|$]]|$1|L|3|M|5|K|7|17|8|@$A|18|B|19|N|O]|$A|1A|B|1B|N|O]|$A|1C|B|1D|N|O]|$A|1E|B|1F|N|O]|$A|1G|B|1H|N|O]|$A|1I|B|1J|N|O]|$A|1K|B|1L|N|O]]|9|@]|C|$]]]|P|$Q|$5|R|S|T|C|$U|V]]|W|$5|R|S|T|C|$U|X]]]]

<h1><a href="https://www.r-project.org/" rel="noreferrer">R</a>, 31 bytes</h1>

<pre class="lang-r prettyprint-override"><code>function(x,y)nchar((x/y)%%1)==3
</code></pre>
<p><a href="https://tio.run/##K/qfoJpZHF@cmpuZklmWWZyZlJMan1SpmmD7P600L7kkMz9Po0KnUjMvOSOxSEOjQr9SU1XVUNPW1vi/mQJWnQomCsoKJUWlqVwWeBSkJeYU41ZhCjPCkpACcxwKjOF2/AcA" rel="noreferrer" title="R – Try It Online">Try it online!</a></p>
<p>Fractional part (<code>%%1</code>) of <code>x/y</code> must be 3 characters: so, <code>0.1</code>, <code>0.2</code> ... <code>0.9</code>, but not <code>0</code> or <code>0.3333</code>.</p>


blocks|key|46235|text|Python+2，23字节|type|header-one|depth|inlineStyleRanges|entityRanges|offset|length|data|46236|从克索的回答保存1字节(用于赏金)。请注意，链接的比较强制n*10%25k+=+0和n%25k+>+0都为真。|unstyled|style|CODE|46237|lambda+n,k:1>n*10%25k|code-block|syntax|javascript|46238|在网上试试！|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/2/|1|https://codegolf.stackexchange.com/a/220283/88546|2|https://codegolf.meta.stackexchange.com/a/18334/88546|3|https://tio.run/##K6gsycjPM/qfZhvzPycxNyklUSFPJ9vK0C5Py9BANdsmTzX7f3lGZk6qgqEVFydQSsFWITOvoLREQ5OLs6AoM69EIU0DKKz5P9pCR8EklgtEmQIpSwhlrqNgHAsA^0|0|8|0|0|T|A|14|7|1|5|1|E|2|2|0|0|0|6|0|6|3^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@$A|13|B|14|1|15]]|C|$]]|$1|D|3|E|5|F|7|16|8|@$A|17|B|18|G|H]|$A|19|B|1A|G|H]]|9|@$A|1B|B|1C|1|1D]|$A|1E|B|1F|1|1G]]|C|$]]|$1|I|3|J|5|K|7|1H|8|@]|9|@]|C|$L|M]]|$1|N|3|O|5|F|7|1I|8|@$A|1J|B|1K|G|H]]|9|@$A|1L|B|1M|1|1N]]|C|$]]]|P|$Q|$5|R|S|T|C|$U|V]]|W|$5|R|S|T|C|$U|X]]|Y|$5|R|S|T|C|$U|Z]]|10|$5|R|S|T|C|$U|11]]]]

<h1><a href="https://docs.python.org/2/" rel="noreferrer">Python 2</a>, 23 bytes</h1>
<p>Saves 1 byte from <a href="https://codegolf.stackexchange.com/a/220283/88546">xnor's answer</a> <em>(for the <a href="https://codegolf.meta.stackexchange.com/a/18334/88546">bounty</a>)</em>. Notice that the chained comparison forces <code>n*10%k = 0</code> and <code>n%k &gt; 0</code> to both be true.</p>

<pre class="lang-python prettyprint-override"><code>lambda n,k:1&gt;n*10%k&lt;n%k
</code></pre>
<p><a href="https://tio.run/##K6gsycjPM/qfZhvzPycxNyklUSFPJ9vK0C5Py9BANdsmTzX7f3lGZk6qgqEVFydQSsFWITOvoLREQ5OLs6AoM69EIU0DKKz5P9pCR8EklgtEmQIpSwhlrqNgHAsA" rel="noreferrer" title="Python 2 – Try It Online">Try it online!</a></p>


blocks|key|45367|text|杰夫特，10+字节数|type|header-one|depth|inlineStyleRanges|entityRanges|offset|length|data|45368|我觉得这是对的，我喝醉了！|unstyled|45369|*A+vV+«UvV|code-block|syntax|javascript|45370|试试看|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/ETHproductions/japt|1|https://en.wikipedia.org/wiki/ISO/IEC_8859-1|2|https://petershaggynoble.github.io/Japt-Interpreter/?v=1.4.6&code=KkEgdlYgq1V2Vg&input=NyAz^0|0|3|0|7|3|1|0|0|0|0|3|2^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@$A|Z|B|10|1|11]|$A|12|B|13|1|14]]|C|$]]|$1|D|3|E|5|F|7|15|8|@]|9|@]|C|$]]|$1|G|3|H|5|I|7|16|8|@]|9|@]|C|$J|K]]|$1|L|3|M|5|F|7|17|8|@]|9|@$A|18|B|19|1|1A]]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]|U|$5|P|Q|R|C|$S|V]]|W|$5|P|Q|R|C|$S|X]]]]

<h1><a href="https://github.com/ETHproductions/japt" rel="noreferrer">Japt</a>, 10 <a href="https://en.wikipedia.org/wiki/ISO/IEC_8859-1" rel="noreferrer">bytes</a></h1>
<p>I think this is right; I'm quite drunk!</p>
<pre><code>*A vV «UvV
</code></pre>
<p><a href="https://petershaggynoble.github.io/Japt-Interpreter/?v=1.4.6&amp;code=KkEgdlYgq1V2Vg&amp;input=NyAz" rel="noreferrer">Try it</a></p>


blocks|key|2068383|text|外壳，8字节|type|header-one|depth|inlineStyleRanges|entityRanges|offset|length|data|2068384|€tḊ10/¹⌉|code-block|syntax|javascript|2068385|在网上试试！|unstyled|2068386|使用与我在其他答案中看到的方法不同的方法，如果我找到一种更好的方法来检查一个数字是否在里面，可能会更短。(任何一个单字节内置的高尔夫语言都可能做到这一点)|2068387|解释|2068388|取两个数字a和b，我们计算：\frac{lcm(a,b)}{a}|style|CODE|2068389|这个值将等于必须乘以a的最小数，以使其可被b整除。由于我们不希望a被b整除，所以我们需要它大于1，并且由于我们希望10*a可以被b整除，所以我们需要它是10的除数。|2068390|€tḊ10/¹⌉
+++++++⌉++++Least+common+multiplier+of+the+two+numbers
+++++/¹+++++divided+by+the+first+number
€+++++++++++Find+its+position+in+the+list:+(returns+0+if+missing)
++Ḋ10++++++++divisors+of+ten:+[1,2,5,10]
+t+++++++++++except+the+first:+[2,5,10]|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/barbuz/Husk|1|https://tio.run/##yygtzv6fG6zhpaSga6eglKqpUez2qKkxt6hc89C2/4@a1pQ83NFlaKB/aOejns7///@bKZhwWSiYclkCsbmCMZBtwmVoYKhgaGAAAA^0|0|2|0|0|0|0|6|1|0|0|0|5|1|7|1|E|I|0|A|1|L|1|W|1|Y|1|1L|4|1S|1|0^^$0|@$1|2|3|4|5|6|7|16|8|@]|9|@$A|17|B|18|1|19]]|C|$]]|$1|D|3|E|5|F|7|1A|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|1B|8|@]|9|@$A|1C|B|1D|1|1E]]|C|$]]|$1|L|3|M|5|K|7|1F|8|@]|9|@]|C|$]]|$1|N|3|O|5|6|7|1G|8|@]|9|@]|C|$]]|$1|P|3|Q|5|K|7|1H|8|@$A|1I|B|1J|R|S]|$A|1K|B|1L|R|S]|$A|1M|B|1N|R|S]]|9|@]|C|$]]|$1|T|3|U|5|K|7|1O|8|@$A|1P|B|1Q|R|S]|$A|1R|B|1S|R|S]|$A|1T|B|1U|R|S]|$A|1V|B|1W|R|S]|$A|1X|B|1Y|R|S]|$A|1Z|B|20|R|S]]|9|@]|C|$]]|$1|V|3|W|5|F|7|21|8|@]|9|@]|C|$G|H]]]|X|$Y|$5|Z|10|11|C|$12|13]]|14|$5|Z|10|11|C|$12|15]]]]

<h1><a href="https://github.com/barbuz/Husk" rel="noreferrer">Husk</a>, 8 bytes</h1>
<pre><code>€tḊ10/¹⌉
</code></pre>
<p><a href="https://tio.run/##yygtzv6fG6zhpaSga6eglKqpUez2qKkxt6hc89C2/4@a1pQ83NFlaKB/aOejns7///@bKZhwWSiYclkCsbmCMZBtwmVoYKhgaGAAAA" rel="noreferrer" title="Husk – Try It Online">Try it online!</a></p>
<p>Uses a different approach from what I have seen in other answers, could be even shorter if I found a better way to check if a number is in [2,5,10]. (Any golfing language with a single-byte builtin for 10 could probably do it)</p>
<h3>Explanation</h3>
<p>Taking 2 numbers <span class="math-container">\$a\$</span> and <span class="math-container">\$b\$</span>, we compute: <span class="math-container">$$\frac{lcm(a,b)}{a}$$</span></p>
<p>This value will be equal to the smallest number you have to multiply <span class="math-container">\$a\$</span> by to make it divisible by <span class="math-container">\$b\$</span>. Since we don't want <span class="math-container">\$a\$</span> to be divisible by <span class="math-container">\$b\$</span> we'll need this to be greater than 1, and since we want <span class="math-container">\$10*a\$</span> to be divisible by <span class="math-container">\$b\$</span> we'll need this to be a divisor of 10. In the end we want the result to be one of [2,5,10].</p>
<pre><code>€tḊ10/¹⌉
       ⌉    Least common multiplier of the two numbers
     /¹     divided by the first number
€           Find its position in the list: (returns 0 if missing)
  Ḋ10        divisors of ten: [1,2,5,10]
 t           except the first: [2,5,10]
</code></pre>


blocks|key|44173|text|JavaScript+(Node.js)，18字节|type|header-one|depth|inlineStyleRanges|entityRanges|offset|length|data|44174|n=>k=>n%25k>0>10*n%25k|code-block|syntax|javascript|44175|在网上试试！|unstyled|44176|-2字节，多亏了亚奥塔|44177|-间接来自Neil或Noodle9或dingledooper的1字节|entityMap|0|LINK|mutability|MUTABLE|url|https://nodejs.org/|1|https://tio.run/##y0osSyxOLsosKNHNy09J/Z9m@z/P1i7b1i5PNdvOwM7QQAvI@J@cn1ecn5Oql5OfrpGmYaGpYaKpaa2gr6@ga6eQlphTXMmFocIUoaKkqLQkA12JJWEl5poaxmj2/AcA|2|https://codegolf.stackexchange.com/users/99986/iota^0|0|K|0|0|0|0|6|1|0|8|3|2|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@$A|11|B|12|1|13]]|C|$]]|$1|D|3|E|5|F|7|14|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|15|8|@]|9|@$A|16|B|17|1|18]]|C|$]]|$1|L|3|M|5|K|7|19|8|@]|9|@$A|1A|B|1B|1|1C]]|C|$]]|$1|N|3|O|5|K|7|1D|8|@]|9|@]|C|$]]]|P|$Q|$5|R|S|T|C|$U|V]]|W|$5|R|S|T|C|$U|X]]|Y|$5|R|S|T|C|$U|Z]]]]

<h1><a href="https://nodejs.org" rel="nofollow noreferrer">JavaScript (Node.js)</a>, 18 bytes</h1>

<pre class="lang-javascript prettyprint-override"><code>n=&gt;k=&gt;n%k&gt;0&gt;10*n%k
</code></pre>
<p><a href="https://tio.run/##y0osSyxOLsosKNHNy09J/Z9m@z/P1i7b1i5PNdvOwM7QQAvI@J@cn1ecn5Oql5OfrpGmYaGpYaKpaa2gr6@ga6eQlphTXMmFocIUoaKkqLQkA12JJWEl5poaxmj2/AcA" rel="nofollow noreferrer" title="JavaScript (Node.js) – Try It Online">Try it online!</a></p>
<p>-2 bytes thanks to <a href="https://codegolf.stackexchange.com/users/99986/iota">iota</a></p>
<p>-1 byte indirectly from Neil or Noodle9 or dingledooper</p>


blocks|key|44220|text|哈斯克尔，30字节|type|header-one|depth|inlineStyleRanges|entityRanges|offset|length|data|44221|n#k=(10*n`mod`k<1)>(n`mod`k<1)|code-block|syntax|javascript|44222|在网上试试！|unstyled|44223|受@xnor+回答启发|unordered-list-item|entityMap|0|LINK|mutability|MUTABLE|url|https://www.haskell.org/|1|https://tio.run/##y0gszk7Nyfn/P08521bD0EArLyE3PyUh28ZQ004Dwf6fm5iZp2CrUFCUmVeiEA1UXBOdp5Mda6MbHW2ho2ASqwOiTIGUJYQy11Ewjo2N/Q8A|2|https://codegolf.stackexchange.com/a/220283/84844^0|0|4|0|0|0|0|6|1|0|7|2|2^^$0|@$1|2|3|4|5|6|7|Z|8|@]|9|@$A|10|B|11|1|12]]|C|$]]|$1|D|3|E|5|F|7|13|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|14|8|@]|9|@$A|15|B|16|1|17]]|C|$]]|$1|L|3|M|5|N|7|18|8|@]|9|@$A|19|B|1A|1|1B]]|C|$]]]|O|$P|$5|Q|R|S|C|$T|U]]|V|$5|Q|R|S|C|$T|W]]|X|$5|Q|R|S|C|$T|Y]]]]

<h1><a href="https://www.haskell.org/" rel="nofollow noreferrer">Haskell</a>, 30 bytes</h1>

<pre class="lang-hs prettyprint-override"><code>n#k=(10*n`mod`k&lt;1)&gt;(n`mod`k&lt;1)
</code></pre>
<p><a href="https://tio.run/##y0gszk7Nyfn/P08521bD0EArLyE3PyUh28ZQ004Dwf6fm5iZp2CrUFCUmVeiEA1UXBOdp5Mda6MbHW2ho2ASqwOiTIGUJYQy11Ewjo2N/Q8A" rel="nofollow noreferrer" title="Haskell – Try It Online">Try it online!</a></p>
<ul>
<li>inspired by @xnor <a href="https://codegolf.stackexchange.com/a/220283/84844">answer</a></li>
</ul>


blocks|key|44420|text|果冻，6字节|type|header-one|depth|inlineStyleRanges|entityRanges|offset|length|data|44421|×⁵ọ¹ȧ%25|code-block|syntax|javascript|44422|在网上试试！|unstyled|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/DennisMitchell/jelly|1|https://tio.run/##y0rNyan8///w9EeNWx/u7j2088Ry1f@H2/UfNa35/z862kLHJFYHSJoCSXMwaaljDCQNDQx0zGNjAQ^0|0|2|0|0|0|0|6|1^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|Z|8|@]|9|@$A|10|B|11|1|12]]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]|S|$5|N|O|P|C|$Q|T]]]]

<h1><a href="https://github.com/DennisMitchell/jelly" rel="nofollow noreferrer">Jelly</a>, 6 bytes</h1>
<pre><code>×⁵ọ¹ȧ%
</code></pre>
<p><a href="https://tio.run/##y0rNyan8///w9EeNWx/u7j2088Ry1f@H2/UfNa35/z862kLHJFYHSJoCSXMwaaljDCQNDQx0zGNjAQ" rel="nofollow noreferrer" title="Jelly – Try It Online">Try it online!</a></p>


blocks|key|2066123|text|J，11字节|type|header-one|depth|inlineStyleRanges|entityRanges|offset|length|data|2066124|3=[#@":@%25~%7C|code-block|syntax|javascript|2066125|在网上试试！|unstyled|2066126|看来我无意中发现了与APL答案几乎完全相同的方法。|2066127|[...%25~%7C左输入[浮点，当左输入被划分为右输入%7C时，将其分为%25~。|unordered-list-item|style|CODE|2066128|#@":@格式，结果为字符串并取其长度。|2066129|3=等于3吗？这只适用于表单0.n的数字，其中n是数字1-9的一个元素。|entityMap|0|LINK|mutability|MUTABLE|url|http://jsoftware.com/|1|https://tio.run/##y/qvpKeepmBrpaCuoKNgoGAFxLp6Cs5BPm7/jW2jlR2UrBxU62r@a3KlJmfkK6inJeYUVyokJxanFqtDhEwU0hQsIExjINOcC6qypKi0JANVqSlCKYhpqfAfAA^0|0|1|0|0|0|0|6|1|0|0|0|7|A|1|P|1|W|2|0|0|5|0|0|2|E|3|R|3^^$0|@$1|2|3|4|5|6|7|15|8|@]|9|@$A|16|B|17|1|18]]|C|$]]|$1|D|3|E|5|F|7|19|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|1A|8|@]|9|@$A|1B|B|1C|1|1D]]|C|$]]|$1|L|3|M|5|K|7|1E|8|@]|9|@]|C|$]]|$1|N|3|O|5|P|7|1F|8|@$A|1G|B|1H|Q|R]|$A|1I|B|1J|Q|R]|$A|1K|B|1L|Q|R]|$A|1M|B|1N|Q|R]]|9|@]|C|$]]|$1|S|3|T|5|P|7|1O|8|@$A|1P|B|1Q|Q|R]]|9|@]|C|$]]|$1|U|3|V|5|P|7|1R|8|@$A|1S|B|1T|Q|R]|$A|1U|B|1V|Q|R]|$A|1W|B|1X|Q|R]]|9|@]|C|$]]]|W|$X|$5|Y|Z|10|C|$11|12]]|13|$5|Y|Z|10|C|$11|14]]]]

<h1><a href="http://jsoftware.com/" rel="nofollow noreferrer">J</a>, 11 bytes</h1>
<pre><code>3=[#@&quot;:@%~|
</code></pre>
<p><a href="https://tio.run/##y/qvpKeepmBrpaCuoKNgoGAFxLp6Cs5BPm7/jW2jlR2UrBxU62r@a3KlJmfkK6inJeYUVyokJxanFqtDhEwU0hQsIExjINOcC6qypKi0JANVqSlCKYhpqfAfAA" rel="nofollow noreferrer" title="J – Try It Online">Try it online!</a></p>
<p>Looks like I stumbled on almost exactly the same method as the APL answer.</p>
<ul>
<li><code>[...%~|</code> The left input <code>[</code> floating point divided into <code>%~</code> the remainder when the left input is divided into the right input <code>|</code>.</li>
<li><code>#@&quot;:@</code> Format that result as a string and take its length.</li>
<li><code>3=</code> Does that equal 3?  This will only be true for numbers of the form <code>0.n</code> where n is an element of the digits <code>1-9</code>.</li>
</ul>


blocks|key|2066349|text|PowerShell，36字节|type|header-one|depth|inlineStyleRanges|entityRanges|offset|length|data|2066350|param($n,$k)!((10*$n)%25$k)-and($n%25$k)|code-block|syntax|javascript|2066351|在网上试试！|unstyled|2066352|受@xnor方法的启发，检查10n是否为k的倍数，但n本身不是|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/TryItOnline/TioSetup/wiki/Powershell|1|https://tio.run/##K8gvTy0qzkjNyfn/vyCxKDFXQyVPRyVbU1FDw9BASyVPUxXI0U3MSwGKg5j///@3@G8CAA^0|0|A|0|0|0|0|6|1|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@$A|X|B|Y|1|Z]]|C|$]]|$1|D|3|E|5|F|7|10|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|11|8|@]|9|@$A|12|B|13|1|14]]|C|$]]|$1|L|3|M|5|K|7|15|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]|U|$5|P|Q|R|C|$S|V]]]]

<h1><a href="https://github.com/TryItOnline/TioSetup/wiki/Powershell" rel="nofollow noreferrer">PowerShell</a>, 36 bytes</h1>

<pre class="lang-powershell prettyprint-override"><code>param($n,$k)!((10*$n)%$k)-and($n%$k)
</code></pre>
<p><a href="https://tio.run/##K8gvTy0qzkjNyfn/vyCxKDFXQyVPRyVbU1FDw9BASyVPUxXI0U3MSwGKg5j///@3@G8CAA" rel="nofollow noreferrer" title="PowerShell – Try It Online">Try it online!</a></p>
<p>Inspired by @xnor method of checking if 10n is multiple of k but n itself not</p>


blocks|key|45701|text|GolfScript，15字节|type|header-one|depth|inlineStyleRanges|entityRanges|offset|length|data|45702|.~@~%25!!@10*@%25!&|code-block|syntax|javascript|45703|在网上试试！|unstyled|45704|不幸的是，由于GolfScript缺乏对浮点值的支持，我无法使用查找小数位的策略。程序只是检查n%25k是真实的，10n+%25k是假的。|45705|另外值得注意的是，我发现更容易将输入作为包含两个空格分隔的整数的字符串，而不是直接将其视为两个整数。|45706|.~@~++++++++++++++++++prepare+two+sets+of+n+and+k
++++%25!!+++++++++++++++check+if+n+%25+k+>+0
+++++++@10*+++++++++++multiply+n+by+10
+++++++++++@%25!++++++++check+if+10n+%25+k+=+0
++++++++++++++&+++++++AND+both+values|entityMap|0|LINK|mutability|MUTABLE|url|http://www.golfscript.com/golfscript/|1|https://tio.run/##S8/PSStOLsosKPn/X6/OoU5VUdHB0EDLQVVR7f9/CwVTAA^0|0|A|0|0|0|0|6|1|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@$A|11|B|12|1|13]]|C|$]]|$1|D|3|E|5|F|7|14|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|15|8|@]|9|@$A|16|B|17|1|18]]|C|$]]|$1|L|3|M|5|K|7|19|8|@]|9|@]|C|$]]|$1|N|3|O|5|K|7|1A|8|@]|9|@]|C|$]]|$1|P|3|Q|5|F|7|1B|8|@]|9|@]|C|$G|H]]]|R|$S|$5|T|U|V|C|$W|X]]|Y|$5|T|U|V|C|$W|Z]]]]

<h1><a href="http://www.golfscript.com/golfscript/" rel="nofollow noreferrer">GolfScript</a>, 15 bytes</h1>
<pre><code>.~@~%!!@10*@%!&amp;
</code></pre>
<p><a href="https://tio.run/##S8/PSStOLsosKPn/X6/OoU5VUdHB0EDLQVVR7f9/CwVTAA" rel="nofollow noreferrer" title="GolfScript – Try It Online">Try it online!</a></p>
<p>Unfortunately, as GolfScript lacks support for floating-point values, I couldn't use the strategy of looking for the decimal place. The program just checks if n % k is truthy and 10n % k is falsy.</p>
<p>Also of note is that I found it easier to take the input as a string containing two space-separated integers instead of taking it as two integers directly.</p>
<pre><code>.~@~                  prepare two sets of n and k
    %!!               check if n % k &gt; 0
       @10*           multiply n by 10
           @%!        check if 10n % k = 0
              &amp;       AND both values
</code></pre>


blocks|key|46012|text|R，28字节|type|header-one|depth|inlineStyleRanges|entityRanges|offset|length|data|46013|function(x,y)!(x*10)%25%25y&x%25%25y|code-block|syntax|javascript|46014|在网上试试！|unstyled|46015|从多米尼克的回答获取的测试线束。|entityMap|0|LINK|mutability|MUTABLE|url|https://www.r-project.org/|1|https://tio.run/##K/qfoJpZHF@cmpuZklmWWZyZlJMan1SpmmD7P600L7kkMz9Po0KnUlNRo0LL0EBTVbVSrQJI/DdTwKpNwURBWaGkqDSVywKPgrTEnGLcKkxhRlgSUmCOQ4Ex3A5DSwMDQxyqDA2AAK7yPwA|2|https://codegolf.stackexchange.com/a/220284/55372^0|0|1|0|0|0|0|6|1|0|1|7|2^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@$A|Z|B|10|1|11]]|C|$]]|$1|D|3|E|5|F|7|12|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|13|8|@]|9|@$A|14|B|15|1|16]]|C|$]]|$1|L|3|M|5|K|7|17|8|@]|9|@$A|18|B|19|1|1A]]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]|U|$5|P|Q|R|C|$S|V]]|W|$5|P|Q|R|C|$S|X]]]]

<h1><a href="https://www.r-project.org/" rel="nofollow noreferrer">R</a>, 28 bytes</h1>

<pre class="lang-r prettyprint-override"><code>function(x,y)!(x*10)%%y&amp;x%%y
</code></pre>
<p><a href="https://tio.run/##K/qfoJpZHF@cmpuZklmWWZyZlJMan1SpmmD7P600L7kkMz9Po0KnUlNRo0LL0EBTVbVSrQJI/DdTwKpNwURBWaGkqDSVywKPgrTEnGLcKkxhRlgSUmCOQ4Ex3A5DSwMDQxyqDA2AAK7yPwA" rel="nofollow noreferrer" title="R – Try It Online">Try it online!</a></p>
<p>Test harness taken from <a href="https://codegolf.stackexchange.com/a/220284/55372">Dominic's answer</a>.</p>


blocks|key|2069120|text|Vyxal+v2.6.0%2B，6字节|type|header-one|depth|inlineStyleRanges|entityRanges|offset|length|data|2069121|ġ/₀KḢc|code-block|syntax|javascript|2069122|将其输入作为k，然后是n。|unstyled|style|CODE|2069123|在网上试试！|2069124|这使用与我的正则答案相同的算法，这与狮子座的回答使用的算法类似。它声称{k\over+gcd(n,k)}+∈+[2,5,10]|2069125|它不使用浮点，因此可以正确处理任意大整数。|2069126|ġ++++++++++++Push+gcd(n,k)
+/+++++++++++Pop+the+above+result,+and+push+k+divided+by+it
++₀K+++++++++Push+the+divisors+of+10,+i.e.+[1,+2,+5,+10]
++++Ḣ++++++++Head+remove+–+drop+first+element,+yielding+[2,+5,+10]
+++++c+++++++Is+the+above+quotient+contained+in+the+above+list?|2069127|Vyxalv2.5.3ω，7字节|2069128|ġ/₀Kḣ„c|2069129|2069130|在网上试试！+-+v2.5.3ω在网上试试！+-最新版本|2069131|首先，我没有使用Ḣ+head-remove元素的原因是，我指的是过时的文档--我对git的克隆被指向main而不是main。|2069132|ġ+++++++++++++Push+gcd(n,k)
+/++++++++++++Pop+the+above+result,+and+push+k+divided+by+it
++₀K++++++++++Push+the+divisors+of+10,+i.e.+[1,+2,+5,+10]
++++ḣ+++++++++Head+extract+–+split+the+above,+yielding+1+and+[2,+5,+10]
+++++„++++++++Rotate+stack+left,+such+that+a=the+above+quotient,+and+b=[2,+5,+10]
++++++c+++++++Is+a+contained+in+b?|2069133|2069134|Ḋ⁰₀*¹Ḋ<|2069135|2069136|在网上试试！+-+v2.5.3ω在网上试试！+-最新版本|2069137|这使用了与xnor的Python回答和许多后续答案相同的逻辑，即断言该n≢0\pmod+k\+\land\+10n≡0\pmod+k。|2069138|Ḋ+++++++++++++Push+1+if+n+is+divisible+by+k,+0+otherwise
+⁰++++++++++++Push+n
++₀+++++++++++Push+10
+++*++++++++++Push+n*10,+popping+both+n+and+10
++++¹+++++++++Push+k
+++++Ḋ++++++++Push+1+if+n*10+is+divisible+by+k,+0+otherwise+(pop+both)
++++++<+++++++Is+the+top+boolean+less+than+the+bottom+boolean?|2069139|Vyxalv2.5.3ω，7个字节，6个元素|2069140|k≈↵*⁰Ḋ¯|2069141|将其输入作为n，然后是k。|2069142|在网上试试！+-+v2.5.3ω在网上试试！+-最新版本|2069143|k≈++++++++++++Push+[0,+1]
++↵+++++++++++Raise+10+to+the+power+of+the+above:+[1,+10]
+++*++++++++++Multiply+the+above+list+by+n
++++⁰+++++++++Push+k
+++++Ḋ++++++++Are+the+list+items+divisible+by+k?+(Result+is+a+list)
++++++¯+++++++Take+deltas+(consecutive+differences)+of+the+list.+Iff+the+two
++++++++++++++elements+of+the+list+were+different,+this+will+be+truthy.|2069144|奇怪的是，¯的行为发生了改变。它过去给[a[0]-a[1],+a[1]-a[2],+…]，现在给[a[1]-a[0],+a[2]-a[1],+…]。这并不会改变输出的真实性，而只是真实结果的标志:旧的⟨-1⟩，新的⟨1⟩。|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/Lyxal/Vyxal|1|https://vyxal.pythonanywhere.com/#WyIiLCIiLCLEoS/igoBL4biiYyIsIiIsIjEwXG4xMjM0NTY3ODkxMjM0NTY4OTUiXQ==|2|https://codegolf.stackexchange.com/a/249412/17216|3|https://codegolf.stackexchange.com/a/220433/17216|4|https://github.com/Vyxal/Vyxal/tree/old|5|https://lyxal.pythonanywhere.com/?flags=&code=%25C4%25A1%252F%25E2%2582%2580K%25E1%25B8%25A3%25E2%2580%259Ec&inputs=10%250A123456789123456895&header=&footer=|6|https://vyxal.pythonanywhere.com/#WyIiLCIiLCLEoS/igoBL4bij4oCeYyIsIiIsIjEwXG4xMjM0NTY3ODkxMjM0NTY4OTUiXQ==|7|8|https://lyxal.pythonanywhere.com/?flags=&code=%25E1%25B8%258A%25E2%2581%25B0%25E2%2582%2580*%25C2%25B9%25E1%25B8%258A%253C&inputs=10%250A123456789123456895&header=&footer=|9|https://vyxal.pythonanywhere.com/#WyIiLCIiLCLhuIrigbDigoAqwrnhuIo8IiwiIiwiMTBcbjEyMzQ1Njc4OTEyMzQ1Njg5NSJd|10|https://codegolf.stackexchange.com/a/220283/17216|11|12|https://lyxal.pythonanywhere.com/?flags=&code=k%25E2%2589%2588%25E2%2586%25B5*%25E2%2581%25B0%25E1%25B8%258A%25C2%25AF&inputs=123456789123456895%250A10&header=&footer=|13|https://vyxal.pythonanywhere.com/#WyIiLCIiLCJr4omI4oa1KuKBsOG4isKvIiwiIiwiMTIzNDU2Nzg5MTIzNDU2ODk1XG4xMCJd^0|0|D|0|0|0|6|1|B|1|0|0|6|1|0|Z|S|6|4|2|I|6|3|0|0|0|0|C|4|0|0|6|1|B|1|0|0|6|5|G|6|6|0|8|1|0|0|0|C|7|0|0|6|1|B|1|0|0|6|8|G|6|9|0|Z|V|5|D|A|0|0|0|C|B|0|0|6|1|B|1|0|0|6|C|G|6|D|0|0|5|1|J|P|1C|P|2S|4|2Z|3^^$0|@$1|2|3|4|5|6|7|2N|8|@]|9|@$A|2O|B|2P|1|2Q]]|C|$]]|$1|D|3|E|5|F|7|2R|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|2S|8|@$A|2T|B|2U|L|M]|$A|2V|B|2W|L|M]]|9|@]|C|$]]|$1|N|3|O|5|K|7|2X|8|@]|9|@$A|2Y|B|2Z|1|30]]|C|$]]|$1|P|3|Q|5|K|7|31|8|@$A|32|B|33|L|M]]|9|@$A|34|B|35|1|36]|$A|37|B|38|1|39]]|C|$]]|$1|R|3|S|5|K|7|3A|8|@]|9|@]|C|$]]|$1|T|3|U|5|F|7|3B|8|@]|9|@]|C|$G|H]]|$1|V|3|W|5|6|7|3C|8|@]|9|@$A|3D|B|3E|1|3F]]|C|$]]|$1|X|3|Y|5|F|7|3G|8|@]|9|@]|C|$G|H]]|$1|Z|3|J|5|K|7|3H|8|@$A|3I|B|3J|L|M]|$A|3K|B|3L|L|M]]|9|@]|C|$]]|$1|10|3|11|5|K|7|3M|8|@]|9|@$A|3N|B|3O|1|3P]|$A|3Q|B|3R|1|3S]]|C|$]]|$1|12|3|13|5|K|7|3T|8|@$A|3U|B|3V|L|M]]|9|@]|C|$]]|$1|14|3|15|5|F|7|3W|8|@]|9|@]|C|$G|H]]|$1|16|3|W|5|6|7|3X|8|@]|9|@$A|3Y|B|3Z|1|40]]|C|$]]|$1|17|3|18|5|F|7|41|8|@]|9|@]|C|$G|H]]|$1|19|3|J|5|K|7|42|8|@$A|43|B|44|L|M]|$A|45|B|46|L|M]]|9|@]|C|$]]|$1|1A|3|1B|5|K|7|47|8|@]|9|@$A|48|B|49|1|4A]|$A|4B|B|4C|1|4D]]|C|$]]|$1|1C|3|1D|5|K|7|4E|8|@$A|4F|B|4G|L|M]]|9|@$A|4H|B|4I|1|4J]]|C|$]]|$1|1E|3|1F|5|F|7|4K|8|@]|9|@]|C|$G|H]]|$1|1G|3|1H|5|6|7|4L|8|@]|9|@$A|4M|B|4N|1|4O]]|C|$]]|$1|1I|3|1J|5|F|7|4P|8|@]|9|@]|C|$G|H]]|$1|1K|3|1L|5|K|7|4Q|8|@$A|4R|B|4S|L|M]|$A|4T|B|4U|L|M]]|9|@]|C|$]]|$1|1M|3|1N|5|K|7|4V|8|@]|9|@$A|4W|B|4X|1|4Y]|$A|4Z|B|50|1|51]]|C|$]]|$1|1O|3|1P|5|F|7|52|8|@]|9|@]|C|$G|H]]|$1|1Q|3|1R|5|K|7|53|8|@$A|54|B|55|L|M]|$A|56|B|57|L|M]|$A|58|B|59|L|M]|$A|5A|B|5B|L|M]|$A|5C|B|5D|L|M]]|9|@]|C|$]]]|1S|$1T|$5|1U|1V|1W|C|$1X|1Y]]|1Z|$5|1U|1V|1W|C|$1X|20]]|21|$5|1U|1V|1W|C|$1X|22]]|23|$5|1U|1V|1W|C|$1X|24]]|25|$5|1U|1V|1W|C|$1X|26]]|27|$5|1U|1V|1W|C|$1X|28]]|29|$5|1U|1V|1W|C|$1X|2A]]|2B|$5|1U|1V|1W|C|$1X|26]]|2C|$5|1U|1V|1W|C|$1X|2D]]|2E|$5|1U|1V|1W|C|$1X|2F]]|2G|$5|1U|1V|1W|C|$1X|2H]]|2I|$5|1U|1V|1W|C|$1X|26]]|2J|$5|1U|1V|1W|C|$1X|2K]]|2L|$5|1U|1V|1W|C|$1X|2M]]]]

<h1><a href="https://github.com/Lyxal/Vyxal" rel="nofollow noreferrer">Vyxal v2.6.0+</a>, 6 bytes</h1>
<pre><code>ġ/₀KḢc
</code></pre>
<p>Takes its input as <span class="math-container">\$k\$</span> followed by <span class="math-container">\$n\$</span>.</p>
<p><a href="https://vyxal.pythonanywhere.com/#WyIiLCIiLCLEoS/igoBL4biiYyIsIiIsIjEwXG4xMjM0NTY3ODkxMjM0NTY4OTUiXQ==" rel="nofollow noreferrer">Try it Online!</a></p>
<p>This uses the same algorithm as my <a href="https://codegolf.stackexchange.com/a/249412/17216">regex answer</a>, which is similar to the algorithm used by <a href="https://codegolf.stackexchange.com/a/220433/17216">Leo's Husk answer</a>. It asserts that <span class="math-container">$${k\over gcd(n,k)} ∈ [2,5,10]$$</span></p>
<p>It doesn't use floating point, and can thus handle arbitrarily large integers correctly.</p>
<pre><code>ġ            Push gcd(n,k)
 /           Pop the above result, and push k divided by it
  ₀K         Push the divisors of 10, i.e. [1, 2, 5, 10]
    Ḣ        Head remove – drop first element, yielding [2, 5, 10]
     c       Is the above quotient contained in the above list?
</code></pre>
<h2><a href="https://github.com/Vyxal/Vyxal/tree/old" rel="nofollow noreferrer">Vyxal v2.5.3ω</a>, 7 bytes</h2>
<pre><code>ġ/₀Kḣ„c
</code></pre>
<p>Takes its input as <span class="math-container">\$k\$</span> followed by <span class="math-container">\$n\$</span>.</p>
<p><a href="https://lyxal.pythonanywhere.com?flags=&amp;code=%C4%A1%2F%E2%82%80K%E1%B8%A3%E2%80%9Ec&amp;inputs=10%0A123456789123456895&amp;header=&amp;footer=" rel="nofollow noreferrer">Try it Online!</a> - v2.5.3ω<br />
<a href="https://vyxal.pythonanywhere.com/#WyIiLCIiLCLEoS/igoBL4bij4oCeYyIsIiIsIjEwXG4xMjM0NTY3ODkxMjM0NTY4OTUiXQ==" rel="nofollow noreferrer">Try it Online!</a> - latest</p>
<p>The reason I didn't use the <code>Ḣ</code> head-remove element in the first place is that I was referring to outdated documentation – my clone of the git repo was pointed at master instead of main.</p>
<pre><code>ġ             Push gcd(n,k)
 /            Pop the above result, and push k divided by it
  ₀K          Push the divisors of 10, i.e. [1, 2, 5, 10]
    ḣ         Head extract – split the above, yielding 1 and [2, 5, 10]
     „        Rotate stack left, such that a=the above quotient, and b=[2, 5, 10]
      c       Is a contained in b?
</code></pre>
<h1><a href="https://github.com/Vyxal/Vyxal/tree/old" rel="nofollow noreferrer">Vyxal v2.5.3ω</a>, 7 bytes</h1>
<pre><code>Ḋ⁰₀*¹Ḋ&lt;
</code></pre>
<p>Takes its input as <span class="math-container">\$k\$</span> followed by <span class="math-container">\$n\$</span>.</p>
<p><a href="https://lyxal.pythonanywhere.com?flags=&amp;code=%E1%B8%8A%E2%81%B0%E2%82%80*%C2%B9%E1%B8%8A%3C&amp;inputs=10%0A123456789123456895&amp;header=&amp;footer=" rel="nofollow noreferrer">Try it Online!</a> - v2.5.3ω<br />
<a href="https://vyxal.pythonanywhere.com/#WyIiLCIiLCLhuIrigbDigoAqwrnhuIo8IiwiIiwiMTBcbjEyMzQ1Njc4OTEyMzQ1Njg5NSJd" rel="nofollow noreferrer">Try it Online!</a> - latest</p>
<p>This uses the same logic as in <a href="https://codegolf.stackexchange.com/a/220283/17216">xnor's Python answer</a> and many subsequent answers, which is to assert that <span class="math-container">\$n≢0\pmod k\ \land\ 10n≡0\pmod k\$</span>.</p>
<pre><code>Ḋ             Push 1 if n is divisible by k, 0 otherwise
 ⁰            Push n
  ₀           Push 10
   *          Push n*10, popping both n and 10
    ¹         Push k
     Ḋ        Push 1 if n*10 is divisible by k, 0 otherwise (pop both)
      &lt;       Is the top boolean less than the bottom boolean?
</code></pre>
<h2><a href="https://github.com/Vyxal/Vyxal/tree/old" rel="nofollow noreferrer">Vyxal v2.5.3ω</a>, 7 bytes, 6 elements</h2>
<pre><code>k≈↵*⁰Ḋ¯
</code></pre>
<p>Takes its input as <span class="math-container">\$n\$</span> followed by <span class="math-container">\$k\$</span>.</p>
<p><a href="https://lyxal.pythonanywhere.com?flags=&amp;code=k%E2%89%88%E2%86%B5*%E2%81%B0%E1%B8%8A%C2%AF&amp;inputs=123456789123456895%0A10&amp;header=&amp;footer=" rel="nofollow noreferrer">Try it Online!</a> - v2.5.3ω<br />
<a href="https://vyxal.pythonanywhere.com/#WyIiLCIiLCJr4omI4oa1KuKBsOG4isKvIiwiIiwiMTIzNDU2Nzg5MTIzNDU2ODk1XG4xMCJd" rel="nofollow noreferrer">Try it Online!</a> - latest</p>
<pre><code>k≈            Push [0, 1]
  ↵           Raise 10 to the power of the above: [1, 10]
   *          Multiply the above list by n
    ⁰         Push k
     Ḋ        Are the list items divisible by k? (Result is a list)
      ¯       Take deltas (consecutive differences) of the list. Iff the two
              elements of the list were different, this will be truthy.
</code></pre>
<p>Oddly, the behavior of <code>¯</code> was changed. It used to give <span class="math-container">\$[a[0]-a[1], a[1]-a[2], …]\$</span> but now gives <span class="math-container">\$[a[1]-a[0], a[2]-a[1], …]\$</span>. This doesn't change the truthiness of the output, just the sign of the truthy result: <code>⟨-1⟩</code> in old, <code>⟨1⟩</code> in new.</p>


blocks|key|44132|text|木炭，9字节|type|header-one|depth|inlineStyleRanges|entityRanges|offset|length|data|44133|⁼¹⌕⮌Ｉ∕ＮＮ.|code-block|syntax|javascript|44134|在网上试试！链接是详细的代码版本。输出一个木炭布尔值，即半白痴的-，如果不是，什么也不输出。解释：|unstyled|style|CODE|44135|++++++Ｎ+++++First+input
+++++∕++++++Divided+by
+++++++Ｎ++++Second+input
++++Ｉ+++++++Cast+to+string
+++⮌++++++++Reversed
++⌕+++++++++Index+of
++++++++.+++Literal+`.`
⁼+++++++++++Equal+to
+¹++++++++++Literal+1
++++++++++++Implicitly+print|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/somebody1234/Charcoal|1|https://tio.run/##S85ILErOT8z5/z@gKDOvRMO1sDQxp1jDUEfBLTMvRSMotSy1qDhVwzmxuETDJbMsMyVVwzOvoLTErzQ3KbVIQ1NHAYULBDoKSnpKQNr6/38LBdP/umU5AA^0|0|2|0|0|0|W|1|0|6|1|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@$A|Z|B|10|1|11]]|C|$]]|$1|D|3|E|5|F|7|12|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|13|8|@$A|14|B|15|L|M]]|9|@$A|16|B|17|1|18]]|C|$]]|$1|N|3|O|5|F|7|19|8|@]|9|@]|C|$G|H]]]|P|$Q|$5|R|S|T|C|$U|V]]|W|$5|R|S|T|C|$U|X]]]]

<h1><a href="https://github.com/somebody1234/Charcoal" rel="nofollow noreferrer">Charcoal</a>, 9 bytes</h1>
<pre><code>⁼¹⌕⮌Ｉ∕ＮＮ.
</code></pre>
<p><a href="https://tio.run/##S85ILErOT8z5/z@gKDOvRMO1sDQxp1jDUEfBLTMvRSMotSy1qDhVwzmxuETDJbMsMyVVwzOvoLTErzQ3KbVIQ1NHAYULBDoKSnpKQNr6/38LBdP/umU5AA" rel="nofollow noreferrer" title="Charcoal – Try It Online">Try it online!</a> Link is to verbose version of code. Outputs a Charcoal boolean, i.e. <code>-</code> for semidivisble, nothing if not. Explanation:</p>
<pre><code>      Ｎ     First input
     ∕      Divided by
       Ｎ    Second input
    Ｉ       Cast to string
   ⮌        Reversed
  ⌕         Index of
        .   Literal `.`
⁼           Equal to
 ¹          Literal 1
            Implicitly print
</code></pre>


blocks|key|45239|text|Vyxal+2.0.0，6字节|type|header-one|depth|inlineStyleRanges|entityRanges|offset|length|data|45240|/Ṫt\.=|code-block|syntax|javascript|45241|在网上试试！|unstyled|45242|检查最后一个字符是否为“。”|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/Lyxal/Vyxal|1|http://lyxal.pythonanywhere.com/?flags=&code=%252F%25E1%25B9%25AAt%255C.%253D&inputs=5%250A8&header=&footer=^0|0|B|0|0|0|0|6|1|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@$A|X|B|Y|1|Z]]|C|$]]|$1|D|3|E|5|F|7|10|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|11|8|@]|9|@$A|12|B|13|1|14]]|C|$]]|$1|L|3|M|5|K|7|15|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]|U|$5|P|Q|R|C|$S|V]]]]

<h1><a href="https://github.com/Lyxal/Vyxal" rel="nofollow noreferrer">Vyxal 2.0.0</a>, 6 bytes</h1>
<pre><code>/Ṫt\.=
</code></pre>
<p><a href="http://lyxal.pythonanywhere.com?flags=&amp;code=%2F%E1%B9%AAt%5C.%3D&amp;inputs=5%0A8&amp;header=&amp;footer=" rel="nofollow noreferrer">Try it Online!</a></p>
<p>Checks if the second last character is &quot;.&quot;</p>


blocks|key|2068027|text|外壳，9字节|type|header-one|depth|inlineStyleRanges|entityRanges|offset|length|data|2068028|&¬%25³*10¹%25|code-block|syntax|javascript|2068029|在网上试试！|unstyled|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/barbuz/Husk|1|https://tio.run/##yygtzv6fq3Fo@aOmxkdtEx61TdL8r3ZojeqhzVqGBod2qv5XSq0oSE0uSU1RyC8tKSgtsVIw1DHQMQRCAwUNMx0TheiSotKSjMpYhbT8nJz8cqDCpEqFkoxUBROF9Myy1DyFktTiEt3kxOLUYs3//6OjTXTMYnWApAWQNIWSlkDSWMc8NhYA^0|0|2|0|0|0|0|6|1^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|Z|8|@]|9|@$A|10|B|11|1|12]]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]|S|$5|N|O|P|C|$Q|T]]]]

<h1><a href="https://github.com/barbuz/Husk" rel="nofollow noreferrer">Husk</a>, 9 bytes</h1>
<pre><code>&amp;¬%³*10¹%
</code></pre>
<p><a href="https://tio.run/##yygtzv6fq3Fo@aOmxkdtEx61TdL8r3ZojeqhzVqGBod2qv5XSq0oSE0uSU1RyC8tKSgtsVIw1DHQMQRCAwUNMx0TheiSotKSjMpYhbT8nJz8cqDCpEqFkoxUBROF9Myy1DyFktTiEt3kxOLUYs3//6OjTXTMYnWApAWQNIWSlkDSWMc8NhYA" rel="nofollow noreferrer" title="Husk – Try It Online">Try it online!</a></p>


blocks|key|2068199|text|K+(ngn/k)，13字节|type|header-one|depth|inlineStyleRanges|entityRanges|offset|length|data|2068200|{>/~y!10+1*x}|code-block|syntax|javascript|2068201|在网上试试！|unstyled|2068202|使用@xnor方法。n是x，k是y。|style|CODE|2068203|10+1*x生成10次x和x的列表|unordered-list-item|2068204|y!+mod每一个都是y的|2068205|~不是它们，即检查y是否均匀地划分了10*x和x|2068206|>/返回真正的iif+y均分10*x，而不是x|entityMap|0|LINK|mutability|MUTABLE|url|https://bitbucket.org/ngn/k|1|https://tio.run/##y9bNS8/7/z/NqtpOv65S0dBAwVCrovY/kAJCAzCuU0hT0FNX0LBQMLG2UDC1tgRicwVjIG1oAGSYaP4HAA|2|https://codegolf.stackexchange.com/a/220283/98547^0|0|9|0|0|0|0|6|1|0|A|1|C|1|E|1|G|1|2|7|2|0|0|6|B|1|D|1|0|0|2|B|1|0|0|1|9|1|I|4|N|1|0|0|2|B|1|E|4|M|1^^$0|@$1|2|3|4|5|6|7|19|8|@]|9|@$A|1A|B|1B|1|1C]]|C|$]]|$1|D|3|E|5|F|7|1D|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|1E|8|@]|9|@$A|1F|B|1G|1|1H]]|C|$]]|$1|L|3|M|5|K|7|1I|8|@$A|1J|B|1K|N|O]|$A|1L|B|1M|N|O]|$A|1N|B|1O|N|O]|$A|1P|B|1Q|N|O]]|9|@$A|1R|B|1S|1|1T]]|C|$]]|$1|P|3|Q|5|R|7|1U|8|@$A|1V|B|1W|N|O]|$A|1X|B|1Y|N|O]|$A|1Z|B|20|N|O]]|9|@]|C|$]]|$1|S|3|T|5|R|7|21|8|@$A|22|B|23|N|O]|$A|24|B|25|N|O]]|9|@]|C|$]]|$1|U|3|V|5|R|7|26|8|@$A|27|B|28|N|O]|$A|29|B|2A|N|O]|$A|2B|B|2C|N|O]|$A|2D|B|2E|N|O]]|9|@]|C|$]]|$1|W|3|X|5|R|7|2F|8|@$A|2G|B|2H|N|O]|$A|2I|B|2J|N|O]|$A|2K|B|2L|N|O]|$A|2M|B|2N|N|O]]|9|@]|C|$]]]|Y|$Z|$5|10|11|12|C|$13|14]]|15|$5|10|11|12|C|$13|16]]|17|$5|10|11|12|C|$13|18]]]]

<h1><a href="https://bitbucket.org/ngn/k" rel="nofollow noreferrer">K (ngn/k)</a>, 13 bytes</h1>
<pre><code>{&gt;/~y!10 1*x}
</code></pre>
<p><a href="https://tio.run/##y9bNS8/7/z/NqtpOv65S0dBAwVCrovY/kAJCAzCuU0hT0FNX0LBQMLG2UDC1tgRicwVjIG1oAGSYaP4HAA" rel="nofollow noreferrer" title="K (ngn/k) – Try It Online">Try it online!</a></p>
<p>Uses <a href="https://codegolf.stackexchange.com/a/220283/98547">@xnor's approach</a>. <code>n</code> is <code>x</code>, and <code>k</code> is <code>y</code>.</p>
<ul>
<li><code>10 1*x</code> generate list of 10 times <code>x</code>, and <code>x</code></li>
<li><code>y!</code> mod each of those by <code>y</code></li>
<li><code>~</code> not them, i.e. check if <code>y</code> evenly divides <code>10*x</code> and <code>x</code></li>
<li><code>&gt;/</code> return true iif <code>y</code> evenly divides <code>10*x</code> but not <code>x</code></li>
</ul>


blocks|key|2069493|text|APL，14字节|type|header-one|depth|inlineStyleRanges|entityRanges|data|2069494|{0≠.=⍵%7C1+10×⍺}|code-block|syntax|javascript|2069495|解释。例如，⍺ = 3和⍵ = 5|unstyled|2069496|1+10×⍺+⍝ Make+two-element+vector:+3 30
⍵%7C+++++⍝ Remanders+of+division+to+5:+3 0
0≠.=+++⍝ Check+if+they+are+zero:+0 1+(no yes)
+++++++⍝ Then+check+if+answers+are+different|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|I|7|O|8|@]|9|@]|A|$]]|$1|J|3|K|5|D|7|P|8|@]|9|@]|A|$E|F]]]|L|$]]

<h1>APL, 14 bytes</h1>
<pre><code>{0≠.=⍵|1 10×⍺}
</code></pre>
<p>Explanations. For example, ⍺ = 3 and ⍵ = 5</p>
<pre><code>1 10×⍺ ⍝ Make two-element vector: 3 30
⍵|     ⍝ Remanders of division to 5: 3 0
0≠.=   ⍝ Check if they are zero: 0 1 (no yes)
       ⍝ Then check if answers are different
</code></pre>


blocks|key|44696|text|JavaScript，22字节|type|header-one|depth|inlineStyleRanges|entityRanges|data|44697|n=>k=>/\..$/.test(n/k)|code-block|syntax|javascript|44698|检查除法后是否正好有一个小数位。|unstyled|44699|在网上试试！|offset|length|entityMap|0|LINK|mutability|MUTABLE|url|https://tio.run/##y0osSyxOLsosKNEts/ifZvs/z9Yu29ZOP0ZPT0VfryS1uEQjTz9b839BUWZeiUaahoWmhommJhcS1xSJa4nKNdfUMNbU/A8A^0|0|0|0|0|6|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|I|7|W|8|@]|9|@]|A|$]]|$1|J|3|K|5|I|7|X|8|@]|9|@$L|Y|M|Z|1|10]]|A|$]]]|N|$O|$5|P|Q|R|A|$S|T]]]]

<h1>JavaScript, 22 bytes</h1>
<pre class="lang-js prettyprint-override"><code>n=&gt;k=&gt;/\..$/.test(n/k)
</code></pre>
<p>Checks that there is exactly one decimal place after division.</p>
<p><a href="https://tio.run/##y0osSyxOLsosKNEts/ifZvs/z9Yu29ZOP0ZPT0VfryS1uEQjTz9b839BUWZeiUaahoWmhommJhcS1xSJa4nKNdfUMNbU/A8A" rel="nofollow noreferrer">Try it online!</a></p>


blocks|key|2066584|text|拉库，16字节|type|header-one|depth|inlineStyleRanges|entityRanges|offset|length|data|2066585|*/(.1&none+1)%25%25*|code-block|syntax|javascript|2066586|在网上试试！|unstyled|2066587|这是一个匿名函数，星号表示这两个参数。主表达式是*+/+X+%25%25+*，它检查由表达式X除以的第一个参数是否可以被第二个参数整除。X这里是数字.1和数字1的无交界处的一个和-交界处.Raku线程连接上的表达式，如果第一个由.1除以的参数可被第二个参数除，而第一个由1除以的参数不能被第二个参数整除，则生成一个真实的值。|style|CODE|2066588|返回值是布尔值的连接，它在布尔上下文中折叠为单个值。幸运的是，这个挑战并没有规定返回值必须是两个不同的值之一，或者我必须添加一个so才能将连接折叠成一个普通布尔值，然后再增加两个字节。|2066589|请注意，我使用的是除法而不是乘法，因为需要额外的空间将第一个函数参数从乘法运算符：*+*(10&none+1)%25%25*中分离出来。.1是Raku中的有理数，而不是浮点数，因此不存在浮点舍入错误的危险。|entityMap|0|LINK|mutability|MUTABLE|url|http://raku.org/|1|https://tio.run/##K0gtyjH7n1upoJamYPtfS19Dz1AtLz8vVcFQU1VV67@1QnFipYJ9mkaNSrymQlp@kYKNhYKJnQ6IMgVRlhDKXMHY7j8A^0|0|2|0|0|0|0|6|1|0|O|A|1S|1|1Y|2|23|1|32|2|3N|1|0|1S|2|0|15|H|1S|2^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@$A|13|B|14|1|15]]|C|$]]|$1|D|3|E|5|F|7|16|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|17|8|@]|9|@$A|18|B|19|1|1A]]|C|$]]|$1|L|3|M|5|K|7|1B|8|@$A|1C|B|1D|N|O]|$A|1E|B|1F|N|O]|$A|1G|B|1H|N|O]|$A|1I|B|1J|N|O]|$A|1K|B|1L|N|O]|$A|1M|B|1N|N|O]]|9|@]|C|$]]|$1|P|3|Q|5|K|7|1O|8|@$A|1P|B|1Q|N|O]]|9|@]|C|$]]|$1|R|3|S|5|K|7|1R|8|@$A|1S|B|1T|N|O]|$A|1U|B|1V|N|O]]|9|@]|C|$]]]|T|$U|$5|V|W|X|C|$Y|Z]]|10|$5|V|W|X|C|$Y|11]]]]

<h1><a href="http://raku.org/" rel="nofollow noreferrer">Raku</a>, 16 bytes</h1>
<pre><code>*/(.1&amp;none 1)%%*
</code></pre>
<p><a href="https://tio.run/##K0gtyjH7n1upoJamYPtfS19Dz1AtLz8vVcFQU1VV67@1QnFipYJ9mkaNSrymQlp@kYKNhYKJnQ6IMgVRlhDKXMHY7j8A" rel="nofollow noreferrer" title="Perl 6 – Try It Online">Try it online!</a></p>
<p>This is an anonymous function where the asterisks represent the two arguments.  The main expression is <code>* / X %% *</code>, which checks that the first argument divided by an expression X is divisible by the second argument.  <code>X</code> here is an and-junction of the number <code>.1</code>, and the none-junction of the number <code>1</code>.  Raku threads the expression over the junction, producing a truthy value if the first argument divided by <code>.1</code> is divisble by the second argument, AND the first argument divided by <code>1</code> is NOT divisible by the second argument.</p>
<p>The return value is a junction of boolean values, which collapses into a single value in a boolean context.  Fortunately, this challenge did not stipulate that the return value must be one of two distinct values, or I'd have to add a <code>so</code> to collapse the junction to a regular boolean, for two more bytes.</p>
<p>Note that I used division instead of multiplication because an extra space would have been required to separate the first function argument from the multiplication operator: <code>* *(10&amp;none 1)%%*</code>.  <code>.1</code> is a rational number in Raku, not a floating-point number, so there's no danger of floating-point rounding error.</p>


blocks|key|2066913|text|Java，35字节|type|header-one|depth|inlineStyleRanges|entityRanges|data|2066914|n->k->(n/k%2B"").matches(".%2B\\.[%5E0]")|code-block|syntax|javascript|2066915|我的JavaScript答案端口。|unstyled|offset|length|2066916|在网上试试！|2066917|Java，20字节|2066918|n->k->n*10%25k<1&n%25k>0|2066919|克索的回答港。|2066920|entityMap|0|LINK|mutability|MUTABLE|url|https://codegolf.stackexchange.com/a/220304/99986|1|https://tio.run/##lY1Ba8JAEIXv@RVLTruVTAUrWmxzKt4EwaNaGDcbu8lmd8lOBBF/e7rF2LO@y/DefPOmwhNmVVH3uvGuJVZFDx1pA2VnJWln4WWR@O5gtGTSYAhshdpeEhY1xIGQ4jg5XbAmLvmGWm2P2z3D9hjEjf3Tl4sHajkUf9zsulWFlkgqZ@Vnb7O8znJuX@tRmgpokOSPCjyF0W4H2@/xPhX94r9wcw6kGnAdgY8vyVheAnpvznwugFQg/ibEU/z0Qf79SX428JM7f02u/S8|2|https://codegolf.stackexchange.com/a/220283/99986|3|https://tio.run/##lY1BC4JAEIXv/oq9FCq4KBUVmqfoFgQeo8Omq6yuu4s7ChL@dtvQOue7DO/NN29K0hGvzKqR1Uo2gErjcQuM47wVKTApsBtaqn1ylqKUE63RlTDxspDRHGsgYEYnWYZqs7QTaJgo7g9EmkI7E/vRWZoDepmLo8neGpqxlACNUX4ahRdXXizcwF9VUbAWqyr2x/DXkPQaaI1lC1iZH8CFnWOiFO/tg4OBarC3jrOI3/3JHxfy@5nffPnBGsY3^0|0|0|2|C|0|0|0|6|1|0|0|0|0|5|2|0|0|6|3^^$0|@$1|2|3|4|5|6|7|17|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|18|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|I|7|19|8|@]|9|@$J|1A|K|1B|1|1C]]|A|$]]|$1|L|3|M|5|I|7|1D|8|@]|9|@$J|1E|K|1F|1|1G]]|A|$]]|$1|N|3|O|5|6|7|1H|8|@]|9|@]|A|$]]|$1|P|3|Q|5|D|7|1I|8|@]|9|@]|A|$E|F]]|$1|R|3|S|5|I|7|1J|8|@]|9|@$J|1K|K|1L|1|1M]]|A|$]]|$1|T|3|M|5|I|7|1N|8|@]|9|@$J|1O|K|1P|1|1Q]]|A|$]]]|U|$V|$5|W|X|Y|A|$Z|10]]|11|$5|W|X|Y|A|$Z|12]]|13|$5|W|X|Y|A|$Z|14]]|15|$5|W|X|Y|A|$Z|16]]]]

<h1>Java, 35 bytes</h1>
<pre class="lang-java prettyprint-override"><code>n-&gt;k-&gt;(n/k+&quot;&quot;).matches(&quot;.+\\.[^0]&quot;)
</code></pre>
<p>Port of my <a href="https://codegolf.stackexchange.com/a/220304/99986">JavaScript answer</a>.</p>
<p><a href="https://tio.run/##lY1Ba8JAEIXv@RVLTruVTAUrWmxzKt4EwaNaGDcbu8lmd8lOBBF/e7rF2LO@y/DefPOmwhNmVVH3uvGuJVZFDx1pA2VnJWln4WWR@O5gtGTSYAhshdpeEhY1xIGQ4jg5XbAmLvmGWm2P2z3D9hjEjf3Tl4sHajkUf9zsulWFlkgqZ@Vnb7O8znJuX@tRmgpokOSPCjyF0W4H2@/xPhX94r9wcw6kGnAdgY8vyVheAnpvznwugFQg/ibEU/z0Qf79SX428JM7f02u/S8" rel="nofollow noreferrer">Try it online!</a></p>
<h1>Java, 20 bytes</h1>
<pre class="lang-java prettyprint-override"><code>n-&gt;k-&gt;n*10%k&lt;1&amp;n%k&gt;0
</code></pre>
<p>Port of <a href="https://codegolf.stackexchange.com/a/220283/99986">xnor's answer</a>.</p>
<p><a href="https://tio.run/##lY1BC4JAEIXv/oq9FCq4KBUVmqfoFgQeo8Omq6yuu4s7ChL@dtvQOue7DO/NN29K0hGvzKqR1Uo2gErjcQuM47wVKTApsBtaqn1ylqKUE63RlTDxspDRHGsgYEYnWYZqs7QTaJgo7g9EmkI7E/vRWZoDepmLo8neGpqxlACNUX4ahRdXXizcwF9VUbAWqyr2x/DXkPQaaI1lC1iZH8CFnWOiFO/tg4OBarC3jrOI3/3JHxfy@5nffPnBGsY3" rel="nofollow noreferrer">Try it online!</a></p>


blocks|key|45574|text|Scala，20字节|type|header-one|depth|inlineStyleRanges|entityRanges|data|45575|n=>k=>n%25k-n*10%25k*k>0|code-block|syntax|javascript|45576|在Scastie里试试|unstyled|offset|length|45577|如果n可被k整除，而不能被n*10整除，则n%25k-n*10%25k*k为负。如果两者都可以被k整除，则为0。如果n*10可以被k整除，那么它就是正的。如果两者都不能被k整除，那么它仍然是负的，因为我们将第二个除以k使其变大。|style|CODE|entityMap|0|LINK|mutability|MUTABLE|url|https://scastie.scala-lang.org/uFovZj1ZReWVY1LcySfBvA^0|0|0|0|B|0|0|2|1|5|1|D|4|L|C|18|1|1I|4|1P|1|29|1|2W|1^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|I|7|Y|8|@]|9|@$J|Z|K|10|1|11]]|A|$]]|$1|L|3|M|5|I|7|12|8|@$J|13|K|14|N|O]|$J|15|K|16|N|O]|$J|17|K|18|N|O]|$J|19|K|1A|N|O]|$J|1B|K|1C|N|O]|$J|1D|K|1E|N|O]|$J|1F|K|1G|N|O]|$J|1H|K|1I|N|O]|$J|1J|K|1K|N|O]]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V]]]]

<h1>Scala, 20 bytes</h1>
<pre class="lang-scala prettyprint-override"><code>n=&gt;k=&gt;n%k-n*10%k*k&gt;0
</code></pre>
<p><a href="https://scastie.scala-lang.org/uFovZj1ZReWVY1LcySfBvA" rel="nofollow noreferrer">Try it in Scastie</a></p>
<p>If <code>n</code> is divisible by <code>k</code> but not <code>n*10</code>, <code>n%k-n*10%k*k</code> is negative. If the both are divisible by <code>k</code>, it's 0. If only <code>n*10</code> is divisible by <code>k</code>, it's positive. If neither is divisible by <code>k</code>, it's still negative, because we're multiplying the second by <code>k</code> to make it bigger.</p>


blocks|key|2067942|text|被认为有害的模板，51字节|type|header-one|depth|inlineStyleRanges|entityRanges|offset|length|data|2067943|Fun,A<1>>,A<2>>,F,Rem,A<2>>>>|code-block|syntax|javascript|2067944|在网上试试！|unstyled|2067945|克索的回答的实现。接受两个参数的匿名函数。|2067946|Fun<
++++If,A<1>>,A<2>>,++#+if+10n%25k+>+0
+++++++F,++++++++++++++++++++++++++#+false
+++++++Rem,A<2>>>>++++++++++++#+else+n%25k|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/feresum/tmp-lang|1|https://tio.run/##K0nNLchJLEkt/u9YYPPfrTTPxjPNJig118a3NMfG08bQwE7H0cbQDkQaAUk3HZAcSAQiYGf3X4eL09PG0g5Mmdpx2f0HAA|2|https://codegolf.stackexchange.com/a/220283/66421^0|0|8|0|0|0|0|6|1|0|0|5|2|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@$A|11|B|12|1|13]]|C|$]]|$1|D|3|E|5|F|7|14|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|15|8|@]|9|@$A|16|B|17|1|18]]|C|$]]|$1|L|3|M|5|K|7|19|8|@]|9|@$A|1A|B|1B|1|1C]]|C|$]]|$1|N|3|O|5|F|7|1D|8|@]|9|@]|C|$G|H]]]|P|$Q|$5|R|S|T|C|$U|V]]|W|$5|R|S|T|C|$U|X]]|Y|$5|R|S|T|C|$U|Z]]]]

<h1><a href="https://github.com/feresum/tmp-lang" rel="nofollow noreferrer">Templates Considered Harmful</a>, 51 bytes</h1>
<pre><code>Fun&lt;If&lt;Rem&lt;Mul&lt;I&lt;10&gt;,A&lt;1&gt;&gt;,A&lt;2&gt;&gt;,F,Rem&lt;A&lt;1&gt;,A&lt;2&gt;&gt;&gt;&gt;
</code></pre>
<p><a href="https://tio.run/##K0nNLchJLEkt/u9YYPPfrTTPxjPNJig118a3NMfG08bQwE7H0cbQDkQaAUk3HZAcSAQiYGf3X4eL09PG0g5Mmdpx2f0HAA" rel="nofollow noreferrer" title="Templates Considered Harmful – Try It Online">Try it online!</a></p>
<p>Implementation of <a href="https://codegolf.stackexchange.com/a/220283/66421">xnor's answer</a>. Anonymous function that takes 2 arguments.</p>
<pre><code>Fun&lt;
    If&lt;Rem&lt;Mul&lt;I&lt;10&gt;,A&lt;1&gt;&gt;,A&lt;2&gt;&gt;,  # if 10n%k &gt; 0
       F,                          # false
       Rem&lt;A&lt;1&gt;,A&lt;2&gt;&gt;&gt;&gt;            # else n%k
</code></pre>


blocks|key|2069550|text|布氏对数，7字节|type|header-one|depth|inlineStyleRanges|entityRanges|offset|length|data|2069551|/ṫkt"."|code-block|syntax|javascript|2069552|在网上试试！|unstyled|2069553|解释|2069554|/++++++++++Divide+n+by+k
+ṫ+++++++++Cast+to+string
++kt"."++++The+penultimate+character+must+be+"."|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/JCumin/Brachylog|1|https://tio.run/##SypKTM6ozMlPN/r/X//hztXZJUp6Sv//R1vomMb@BwA^0|0|4|0|0|0|0|6|1|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@$A|Z|B|10|1|11]]|C|$]]|$1|D|3|E|5|F|7|12|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|13|8|@]|9|@$A|14|B|15|1|16]]|C|$]]|$1|L|3|M|5|6|7|17|8|@]|9|@]|C|$]]|$1|N|3|O|5|F|7|18|8|@]|9|@]|C|$G|H]]]|P|$Q|$5|R|S|T|C|$U|V]]|W|$5|R|S|T|C|$U|X]]]]

<h1><a href="https://github.com/JCumin/Brachylog" rel="nofollow noreferrer">Brachylog</a>, 7 bytes</h1>
<pre><code>/ṫkt&quot;.&quot;
</code></pre>
<p><a href="https://tio.run/##SypKTM6ozMlPN/r/X//hztXZJUp6Sv//R1vomMb@BwA" rel="nofollow noreferrer" title="Brachylog – Try It Online">Try it online!</a></p>
<h3>Explanation</h3>
<pre><code>/          Divide n by k
 ṫ         Cast to string
  kt&quot;.&quot;    The penultimate character must be &quot;.&quot;
</code></pre>


blocks|key|46357|text|Dyalog+Unicode，18字节|type|header-one|depth|inlineStyleRanges|entityRanges|offset|length|data|46358|{≠/0=1%7Cq,10×q←⍺÷⍵}|code-block|syntax|javascript|46359|不像这个解决方案那么聪明，但是你会怎么做.*耸耸肩：|unstyled|46360|≠/+⍝+fold+by+way+of+not+equals+(xor)
++0=1%7C+⍝+divisibility+check
++++++q,10×q+⍝+concatenate+
+++++++++++q←⍺÷⍵+⍝+assignment|46361|在网上试试！|entityMap|0|LINK|mutability|MUTABLE|url|https://www.dyalog.com/|1|https://codegolf.stackexchange.com/a/220285/92642|2|https://tio.run/##SyzI0U2pTMzJT///P@1R24TqR50L9A1sDWsKdQwNDk8vBAo96t11ePuj3q21//9bKKQpmHCBSFMuSzBpDiSNAQ^0|0|E|0|0|0|4|4|1|0|0|0|6|2^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@$A|11|B|12|1|13]]|C|$]]|$1|D|3|E|5|F|7|14|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|15|8|@]|9|@$A|16|B|17|1|18]]|C|$]]|$1|L|3|M|5|F|7|19|8|@]|9|@]|C|$G|H]]|$1|N|3|O|5|K|7|1A|8|@]|9|@$A|1B|B|1C|1|1D]]|C|$]]]|P|$Q|$5|R|S|T|C|$U|V]]|W|$5|R|S|T|C|$U|X]]|Y|$5|R|S|T|C|$U|Z]]]]

<h1><a href="https://www.dyalog.com/" rel="nofollow noreferrer">APL (Dyalog Unicode)</a>, 18 bytes</h1>

<pre class="lang-apl prettyprint-override"><code>{≠/0=1|q,10×q←⍺÷⍵}
</code></pre>
<p>Not as clever as this <a href="https://codegolf.stackexchange.com/a/220285/92642">solution</a>, but whatcha gonna do... :shrug:</p>
<pre class="lang-apl prettyprint-override"><code>≠/ ⍝ fold by way of not equals (xor)
  0=1| ⍝ divisibility check
      q,10×q ⍝ concatenate 
           q←⍺÷⍵ ⍝ assignment
</code></pre>
<p><a href="https://tio.run/##SyzI0U2pTMzJT///P@1R24TqR50L9A1sDWsKdQwNDk8vBAo96t11ePuj3q21//9bKKQpmHCBSFMuSzBpDiSNAQ" rel="nofollow noreferrer" title="APL (Dyalog Unicode) – Try It Online">Try it online!</a></p>


blocks|key|46461|text|VBScript，45字节|type|header-one|depth|inlineStyleRanges|entityRanges|data|46462|sub+f(n,k)
msgbox+10*n+mod+k<>n+mod+k
end+sub|code-block|syntax|javascript|46463|移植@xnor的答案有点改变了。|unstyled|46464|检查10n是否为k的倍数，但不检查这些条件是否与<>不相等|offset|length|style|CODE|entityMap^0|0|0|0|O|2^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|I|7|S|8|@]|9|@]|A|$]]|$1|J|3|K|5|I|7|T|8|@$L|U|M|V|N|O]]|9|@]|A|$]]]|P|$]]

<h1>VBScript, 45 bytes</h1>
<pre><code>sub f(n,k)
msgbox 10*n mod k&lt;&gt;n mod k
end sub
</code></pre>
<p>Porting of @xnor's answer a little changed.</p>
<p>Check if 10n is multiple of k but n itself not by checking those conditions aren't equal with <code>&lt;&gt;</code></p>


blocks|key|47603|text|皮耶特+%2B+ascii-piet，37个字节(29×2=58代码)|type|header-one|depth|inlineStyleRanges|entityRanges|offset|length|data|47604|TABLiIiIAQJjJjJBDFjJBRLBiIIAQJAMVcRrr|code-block|syntax|javascript|47605|试试Piet在线吧！|unstyled|47606|作为一幅图像：|47607|📷|atomic|47608|这只是旋转，使ASCII编码更短。在ASCII中，不被轮换，不受限制：|47609|tabliiaqjjjbdfjbrlbiiaqjamvcr
++++ii+++jj+++j++++i+++++++rr|47610|47611|47612|由布布勒的解释器生成的解释：|47613|47614|所以基本上输入，重复，输入，重复，推栈三个值的顶值向下(产生b，a，a，b)，乘以10，交换，模，逻辑不，推堆栈的两个值的顶值向下，模，乘法，逻辑不，逻辑不，输出。|entityMap|0|LINK|mutability|MUTABLE|url|https://www.dangermouse.net/esoteric/piet.html|1|https://github.com/dloscutoff/ascii-piet|2|https://bubbler-4.github.io/piet/#eJx1jsEKwyAQRH8l7HkFNaZVf8V4KKmFQGoDNqfSf-_SrsFS6sE36ww7PmC6nRP4ECQqFzGoA9N8WEeyZUt-55StsPtdFyjN7NtMdfV3i_ltc3_b9q8qJmNA_d6sY0QoaQUPQghAWOYraSXp0JTKBP5yWkpCmPO63ckznR0zhcc8NMqx6rsjPF_CRkux|3|IMAGE|IMMUTABLE|imageUrl|https://i.stack.imgur.com/DD64n.png|imageAlt|4|https://i.stack.imgur.com/5o8vN.png|5|https://i.stack.imgur.com/UjcxV.png^0|0|3|0|6|A|1|0|0|0|A|2|0|0|0|1|3|0|0|0|0|0|1|4|0|0|0|1|5|0^^$0|@$1|2|3|4|5|6|7|1M|8|@]|9|@$A|1N|B|1O|1|1P]|$A|1Q|B|1R|1|1S]]|C|$]]|$1|D|3|E|5|F|7|1T|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|1U|8|@]|9|@$A|1V|B|1W|1|1X]]|C|$]]|$1|L|3|M|5|K|7|1Y|8|@]|9|@]|C|$]]|$1|N|3|O|5|P|7|1Z|8|@]|9|@$A|20|B|21|1|22]]|C|$]]|$1|Q|3|R|5|K|7|23|8|@]|9|@]|C|$]]|$1|S|3|T|5|F|7|24|8|@]|9|@]|C|$G|H]]|$1|U|3|M|5|K|7|25|8|@]|9|@]|C|$]]|$1|V|3|O|5|P|7|26|8|@]|9|@$A|27|B|28|1|29]]|C|$]]|$1|W|3|X|5|K|7|2A|8|@]|9|@]|C|$]]|$1|Y|3|O|5|P|7|2B|8|@]|9|@$A|2C|B|2D|1|2E]]|C|$]]|$1|Z|3|10|5|K|7|2F|8|@]|9|@]|C|$]]]|11|$12|$5|13|14|15|C|$16|17]]|18|$5|13|14|15|C|$16|19]]|1A|$5|13|14|15|C|$16|1B]]|1C|$5|1D|14|1E|C|$1F|1G|1H|-4]]|1I|$5|1D|14|1E|C|$1F|1J|1H|-4]]|1K|$5|1D|14|1E|C|$1F|1L|1H|-4]]]]

<h1><a href="https://www.dangermouse.net/esoteric/piet.html" rel="nofollow noreferrer">Piet</a> + <a href="https://github.com/dloscutoff/ascii-piet" rel="nofollow noreferrer">ascii-piet</a>, 37 bytes (29×2=58 codels)</h1>
<pre class="lang-none prettyprint-override"><code>TABLiIiIAQJjJjJBDFjJBRLBiIIAQJAMVcRrr
</code></pre>
<p><a href="https://bubbler-4.github.io/piet/#eJx1jsEKwyAQRH8l7HkFNaZVf8V4KKmFQGoDNqfSf-_SrsFS6sE36ww7PmC6nRP4ECQqFzGoA9N8WEeyZUt-55StsPtdFyjN7NtMdfV3i_ltc3_b9q8qJmNA_d6sY0QoaQUPQghAWOYraSXp0JTKBP5yWkpCmPO63ckznR0zhcc8NMqx6rsjPF_CRkux" rel="nofollow noreferrer">Try Piet online!</a></p>
<p>As an image:</p>
<p><a href="https://i.stack.imgur.com/DD64n.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DD64n.png" alt="enter image description here" /></a></p>
<p>This is just rotated to make the ASCII encoding shorter. In ASCII, unminified, without being rotated:</p>
<pre><code>tabliiaqjjjbdfjbrlbiiaqjamvcr
    ii   jj   j    i       rr
</code></pre>
<p>As an image:</p>
<p><a href="https://i.stack.imgur.com/5o8vN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5o8vN.png" alt="enter image description here" /></a></p>
<p>Explanation generated by Bubbler's interpreter:</p>
<p><a href="https://i.stack.imgur.com/UjcxV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UjcxV.png" alt="enter image description here" /></a></p>
<p>So basically input, duplicate, input, duplicate, push top value of stack three values down (producing b,a,a,b), multiply by 10, swap, modulo, logical not, push top value of stack two values down, modulo, multiply, logical not, logical not, output.</p>


<p><strong>NOTE: Some terminology used in this challenge is fake.</strong></p>
<p>For two integers <code>n</code> and <code>k</code> both greater than or equal to 2 with <code>n &gt; k</code>, <code>n</code> is <em>semidivisible</em> by <code>k</code> if and only if <code>n/k = r/10</code> for some integer <code>r</code>. However, <code>n</code> may not be divisible by <code>k</code>. Put more simply, the base 10 representation of <code>n/k</code> has exactly one digit after the decimal place. For example, 6 is semidivisible by 4 because 6/4=15/10, but 8 is not semidivisible by 4 because <code>8 % 4 == 0</code>.</p>
<p>Your task is to write a program which takes in two integers as input, in any convenient format, and outputs a truthy (respectively falsy) value if the first input is semidivisible by the second, and a falsey (respectively truthy) value otherwise. Standard loopholes are forbidden. You may assume that <code>n &gt; k</code> and that both <code>n</code> and <code>k</code> are at least 2.</p>
<p>Test cases:</p>
<pre><code>[8, 4] -&gt; falsey
[8, 5] -&gt; truthy
[9, 5] -&gt; truthy
[7, 3] -&gt; falsey
</code></pre>
<p>This question is <a href="/questions/tagged/code-golf" class="post-tag" title="show questions tagged &#39;code-golf&#39;" rel="tag">code-golf</a> therefore shortest answer in bytes wins.</p>


Semidivisibility

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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NOTE:在这个挑战中使用的一些术语是假的。对于带有n的两个整数k和大于或等于2的整数，n是由k半整除的当且仅当某些整数r的n/k = r/10。然而，n可能不能被k整除。更简单地说，n/k的基数10表示在小数位之后正好有一个数字。例如，6可以被4整除，因为6/4=15/10，而8不能被4整除，因为8 % 4 == 0。您的任务是编写一个程序，该程序以任何方便的格式接受两个整数作为输入，如果第一个

问半整除性
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回答 27

Code Golf用户

Python 2，24字节

Code Golf用户

R，31字节

Code Golf用户

Python 2，23字节

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