Question 849902
Digits a, b, and c can be chosen to make the following 
multiplication work.  What is the 3-digit number <i>abc</i>.
<pre><b>
 
        a b c
<u>×         2 4</u>
1  c  b  a  2

{{{24(100a+10b+c)}}}{{{""=""}}}{{{10000+1000*c+100*b+10*a+2}}}

It's obvious that c must be either 8 or 3, since 8x4=32 and 4x3=12,
the only ways to get a last digit of 2.

We try c=8

{{{24(100a+10b+8)}}}{{{""=""}}}{{{10000+1000*8+100*b+10*a+2}}}

Simplify

{{{2400*a+240*b+192)}}}{{{""=""}}}{{{10000+1000*8+100*b+10*a+2}}}

{{{2400*a+240*b+192)}}}{{{""=""}}}{{{10000+8000+100*b+10*a+2}}}

{{{2400*a+240*b+192)}}}{{{""=""}}}{{{18000+100*b+10*a+2}}}

{{{2400*a+240*b+192)}}}{{{""=""}}}{{{18002+100*b+10*a}}}

{{{2390*a+140*b}}}{{{""=""}}}{{{17810}}}

Divide through by 10

{{{239*a+14*b}}}{{{""=""}}}{{{1781}}}

Even if a and b were both the greatest possible digits, which
is 9, the left side would be only 23*9+14*9 = 333, far short
of 1781.  Thus we have ruled out c=8 as possible.  So c=3

{{{24(100a+10b+3)}}}{{{""=""}}}{{{10000+1000*3+100*b+10*a+2}}}

Simplify:

{{{2400*a+240*b+72)}}}{{{""=""}}}{{{10000+1000*3+100*b+10*a+2}}}

{{{2400*a+240*b+72)}}}{{{""=""}}}{{{10000+3000+100*b+10*a+2}}}

{{{2400*a+240*b+72)}}}{{{""=""}}}{{{1300+100*b+10*a+2}}}

{{{2400*a+240*b+72)}}}{{{""=""}}}{{{13002+100*b+10*a}}}

{{{2390*a+140*b}}}{{{""=""}}}{{{12930}}}

Divide through by 10

{{{239a+14b}}}{{{""=""}}}{{{1293}}}

We solve this Diophantine equation:

Since 14 is the coefficient with the smallest
absolute value, 14, we write the 239 and the
1293 in terms of their closest multiple of 14.

239/14 = 17.071...

So the closest multiple of 14 to 239 is 14*17=238
and so 239 = 238+1

1293/14 = 92.357...

So the closest multiple of 14 to 1293 is 14*92=1288
and so 1293 = 1288+5

So

{{{239a+14b}}}{{{""=""}}}{{{1293}}}

becomes

{{{(238+1)a+14b}}}{{{""=""}}}{{{1288+5}}}

{{{238a+a+14b}}}{{{""=""}}}{{{1288+5}}}

Divide every term through by 14

{{{238a/14+a+14b/14}}}{{{""=""}}}{{{1288/14+5/14}}}

{{{17a+a/14+b}}}{{{""=""}}}{{{92+5/14}}}

Get all the fractions on the left side,
and other terms on the right side:

{{{a/14-5/14}}}{{{""=""}}}{{{92-17a-b}}}

The right side equals to an integer, so the left side
must also equal to that same integer.  Let that 
integer be N, so:

{{{a/14-5/14}}}{{{""=""}}}{{{N}}} and {{{92-17a-b}}}{{{""=""}}}{{{N}}}

{{{a-5}}}{{{""=""}}}{{{14N}}}

{{{a}}}{{{""=""}}}{{{14N+5}}}

Substitute in 

{{{92-17(14N+5)-b}}}{{{""=""}}}{{{N}}}

{{{92-238N-85-b}}}{{{""=""}}}{{{N}}}

{{{7-238N-b}}}{{{""=""}}}{{{N}}}

{{{7-239N}}}{{{""=""}}}{{{b}}}

Since b is a digit, the only integer
N can be is 0. So N=0 and

{{{7-239*0}}}{{{""=""}}}{{{b}}}

{{{7-0}}}{{{""=""}}}{{{b}}}

{{{7}}}{{{""=""}}}{{{b}}}

Substituting N=0 and b=7 in

{{{92-17a-b}}}{{{""=""}}}{{{N}}}

{{{92-17a-7}}}{{{""=""}}}{{{0}}}

{{{85-17a}}}{{{""=""}}}{{{0}}}

{{{85}}}{{{""=""}}}{{{17a}}}

{{{85/17}}}{{{""=""}}}{{{17a/17}}}

{{{5=a}}}

So a=5, b=7 and c=3

and the three digit number <i>abc</i> is 573

        5 7 3
<u>×         2 4</u>
1  3  7  5  2

Edwin</pre>