<PRE><B>Question 1208625</B>
&lt;br&gt;
The value of the 3-digit number ABC is 100A+10B+C.
The value of the 2-digit number AC is 10A+C.&lt;br&gt;
The 3-digit number is 16 times the 2-digit number:&lt;br&gt;
{{{100A+10B+C=16(10A+C)}}}
{{{100A+10B+C=160A+16C}}}
{{{10B=60A+15C}}}
{{{B=6A+1.5C}}} [1]&lt;br&gt;
B is a single-digit integer; A is a single-digit integer so 6A is an integer.  That means 1.5C must be an integer, which means C is an even single-digit integer.&lt;br&gt;
Try different values of C in equation [1] to find which ones give single-digit values for B:&lt;br&gt;
C=0: B=6A so A=1 and B=6; ABC is 160.  160/16 = 10 so that solution is good&lt;br&gt;
C=2: B=6A+3 so A is 1 and B is 9; ABC is 192.  192/16 = 12 so that solution is good&lt;br&gt;
For larger values of C, B=6A+1.5C will make B no longer a single-digit integer, so there are no more solutions.&lt;br&gt;
The two numbers ABC that satisfy the given condition are 160 and 192.&lt;br&gt;
ANSWER: 192&lt;br&gt;
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