Question 1201159
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his PIN code is ABCD; and multiplied by 4 it is DCBA.<br><pre>
  ABCD
  x  4
  ----
  DCBA</pre>
4 is an even digit; an even digit times any digit yields an even last digit, so A is even.<br>
4 times ABCD is still a 4-digit number; then, since A must be even, A can only be 2.<br><pre>
  2BCD
  x  4
  ----
  DCB2</pre>
The multiplication D times 4 yields final digit 2, so D can be only 3 or 8.  But 4 times 2BCD yields first digit D, so D must be 8.<br><pre>
  2BC8
  x  4
  ----
  8CB2</pre>
8CB2 is a multiple of 4, so the last 2 digits of 8CB2 must be a multiple of 4.  That means B is odd.<br>
But 4 times 2BC8 yields a 4-digit product, so B must be 1.<br><pre>
  21C8
  x  4
  ----
  8C12</pre>
4 times the last 2 digits C8 yields last 2 digits 12 in the product; that means C can be only 2 or 7.  But since 4 times 21C8 yields the product 8C12, C must be 7.<br><pre>
  2178
  x  4
  ----
  8712</pre>
ANSWER: His PIN code is 2178<br>