Question 126252



Let {{{M}}} be the two integer that are respectively 13 times
the sum of its digits.

Note that {{{M}}} {{{MUST}}}be {{{three-digit}}} number (the digits are from 0-9, and 9*13=117).

 Let {{{M = 100a +10b+c}}}

Then {{{13(a + b+c) = 100a +10b+c  }}}

We can show that M cannot be a{{{ 4-digit}}} number. 
If M is the number {{{4-digit}}} number {{{abcd}}} and all digits are {{{9s}}}, then

{{{13 (a + b + c + d) <= 13 * 9 * 4 = 468 < 1000}}}….=>… not possible 

Similarly, larger numbers are not possible.

We can also show that {{{M }}}cannot be a {{{2-digit}}} number and both digits are {{{9s}}}. 

If {{{M}}} is the number{{{ ab}}}, then

{{{13 (a + b ) = 10a+b  }}}….

=>… {{{13 a + 13b  = 10a+b  }}}…

.=>… {{{13 a – 10a + 13b - b  = 0}}}…

.=>… {{{3 a +12b  = 0}}}….=>… 

 {{{a + 4b  = 0}}}

.=>…{{{not}}} possible since {{{a > 0}}} and {{{b >= 0}}}
  
So the answers can only be {{{3-digit}}} numbers. Let{{{ M}}} be {{{abc}}}, then

{{{13(a + b+c) = 100a +10b+c }}} 
	
{{{13a + 13b +13c = 100a + 10b + c}}}

{{{0 =87a -3b -12c}}}……………..divide by {{{3}}}

{{{87a/3 =3b /3 +12c/3 }}}

{{{29a=b  + 4c}}}

If {{{a = 1}}}, the equation becomes {{{29 = b + 4c}}}. We have the following sets of solutions:

{{{a}}} | {{{b}}} | {{{c}}}
{{{1}}} |{{{1}}} |  {{{7}}}
{{{1}}} | {{{5}}} | {{{6}}}
{{{1}}} | {{{9}}} | {{{5}}}

If {{{a = 2}}}, the equation becomes {{{58 = b + 4c}}}. This has no solution as the largest value of {{{b + 4c}}} is {{{45}}} (when {{{b = 9}}} and 
{{{c = 9}}}). 

Similarly, there is no solution for {{{a>2}}}

Therefore, the possible values of {{{M }}}are {{{117}}}, {{{156}}} and {{{195}}}.