Question 1204315

write a three-digit number {{{ABC}}} as:

{{{ABC = 100 A + 10 B + C}}}

and the number {{{AC}}} (divisor) is:

{{{AC = 10 A + C}}}

We know that {{{(ABC) / (AC )= 11}}} with no remainder,
 
{{{ABC = 11AC }}}

or

{{{100A + 10B + C = 11 (10A + C)}}}

{{{100 A + 10 B + C = 110 A + 11C}}}

{{{10 B   = 110 A -100 A + 11C-C}}}

{{{10 B = 10 A + 10C}}}

Remember that {{{A}}}, {{{B}}}, {{{C}}}, being digits of an integer number, must be whole numbers between {{{0 }}}and {{{9}}}, and that {{{C}}} isn’t zero. 
{{{A }}}must be not {{{0}}} as well, otherwise {{{ABC }}}would have only two digits and not three.

as you can see, {{{10B}}},{{{10 A}}}, and {{{10 C}}} are a multiple of {{{10}}}. 

It’s easy to verify that the only value of {{{C}}} that satisfies this condition is {{{5}}}; all other values for {{{C }}}between {{{1}}} and {{{10 }}} yield a value for {{{10C}}} that is not divisible by {{{10}}}. 

Thus, if {{{C = 5}}}, we can replace {{{5 }}}for {{{C}}} and write:

{{{10 B = 10 A + 50}}}

dividing both terms by {{{10}}}:

{{{B =  A + 5}}}

The only values that satisfy that condition are:

{{{A = 4}}}

{{{B = 4 + 5 = 9}}}

. Let’s summarize:

{{{A = 4}}}

{{{B = 9}}}

{{{C = 5}}}

then

{{{ABC = 495}}}

{{{AC = 45}}}

check:

{{{ 495/45=11}}}


In fact, {{{495/45 = 11 }}} with no remainder.