Question 1085200

let your number be:

{{{abcde}}}

given:

{{{40 000< abcde<70 000}}}

ones digit is two more than ten thousands digit: {{{e=a+2}}}
All my other digits are the same:->{{{b=c=d}}}

The sum of my digit is {{{19}}}:

 {{{a+b+c+d+e=19}}}.....since {{{e=a+2}}}, {{{b=c}}} and {{{d=c}}}, we have

{{{a+c+c+c+a+2=19}}}

{{{2a+3c=19-2}}}

{{{2a+3c=17}}}.........since greater than {{{40000}}} but less than {{{70 000}}}->ten thousands digit could be from {{{4}}} to {{{7}}} but less than {{{7}}}; so, could be {{{a=4}}}, {{{a=5}}} or {{{a=6}}}

since given that ones digit is two more than ten thousands digit: {{{e=a+2}}}-> {{{a}}} cannot be {{{6}}}; so,{{{e}}} must be {{{6}}}, and {{{a=4}}}

{{{2*4+3c=17}}}

{{{8+3c=17}}}

{{{3c=17-8}}}

{{{3c=9}}}

{{{c=3}}} ->then {{{b=3}}} and {{{d=3}}}

and your number is: {{{43336}}}