Question 114750

{{{12A3B}}}

Apply following Divisibility rules: 

Dividing By {{{4}}}:  A number is divisible by {{{4}}} if the {{{last… 2-digits=3B }}} are divisible by {{{4}}}.

 So, digits could be an even number:

{{{B=2}}} or {{{B=6}}} 

Let’s try {{{B=2}}}:
{{{3B= 32}}}
{{{32/4=8}}}………… divisible by {{{4}}}…..

Let’s try {{{B=6}}}:
{{{3B= 36}}}
{{{36/4=9}}}……………… divisible by {{{4}}}…..




Apply now rule Dividing By {{{9}}}:  A number is divisible by{{{ 9}}} if the {{{sum _of_ the _digits}}} is divisible by {{{9}}}; so, {{{sum = 1+2+A+3+B}}}has to be divisible by {{{9}}}.

Now check which {{{B}}} would be right solution:

	{{{12A3B}}}
if {{{B=2}}}, we have {{{12A32}}}


Since the sum of known digits is {{{1+2+ A +3+2 = 8 +A}}}, we need a such of number {{{A}}} to get the sum of the digits as a multiple of {{{9}}} in order 
to get a number {{{12A32}}} divisible by {{{9}}}.
	
 The first number added to this{{{8}}} is {{{1}}}; so,
{{{A=1}}}

if {{{B=6}}}, we have {{{12A36}}}

we see that {{{1+2+A+3+6=12+A}}}, so if {{{B=6}}} then{{{ A=6}}} too because {{{12+6=18}}} and {{{18 }}}is divisible by {{{9}}}; but, to have {{{A=B}}} is not allowed.

So, if {{{A=1}}}and{{{B=2}}} our number is: {{{12132}}}

check


the {{{last… 2-digits=32 }}}…. {{{32}}} is divisible by {{{4}}}

the {{{sum _of_ the _digits = 1+2+1+3+2=9}}}………{{{9}}} is divisible by {{{9}}}

the {{{12132}}}:

{{{12132/4=3033}}}

{{{12132/9=1348}}}