Find the smallest positive integer which ends in n, is divisible by n and whose digits sum to n It's all in the title...
Take as input a positive integer n&gt;=12 and... do what the title says.
Yes, this is on OEIS A187924.
Some test cases
12 -&gt; 912  
13 -&gt; 11713  
14 -&gt; 6314  
15 -&gt; 915  
16 -&gt; 3616  
17 -&gt; 15317  
18 -&gt; 918  
19 -&gt; 17119 
20 -&gt; 9920  
40 -&gt; 1999840   
100-&gt; 99999999999100

This is code-golf. Shortest code in bytes wins!
Befunge, 81 bytes
&amp;&gt;00p0v&lt;!%g0&lt;
v%&quot;d&quot;:_&gt;1+:0^
&gt;00g-&gt;#^_:0v
0g10g-#^_.@1&gt;
&gt;0p:55+/\:v&gt;
^1+g01%+55_$^

Try it online!
Can handle up to n = 70 at least, after which some values will start overflowing the stack cell size on most implementations, and on those that don't, it'll take so long that it's not worth waiting to find out.
Given those constraints, we don't even bother trying to handle values of n greater than 99, which means we can more easily test if the value ends in n with by simply comparing the value modulo 100 with n.
Below is more detailed breakdown of the code.

 Read n from stdin and save in memory.
 Initialize the test value v to 0 and start the main loop, incrementing v up front.
 Test if v%n == 0, and if not return to the start of the main loop.
 Test if v%100 == n, and if not return to the start of the main loop.
 Sum the digits in v by repeatedly adding v modulo 10 and dividing v by 10.
 Test if the sum is equal to n, and if not return to the start of the main loop.
 Otherwise output v and exit.