Question

A four digit number is $202$ times the sum of digits. Altering the thousands with hundreds digit and tens with units and adding this no. with original gives $3333$. Thousands digit is same as tens digit units as hundreds. Find the number

Answer 1

Let us day we have a four digit number $"abcd"$ where $a,b,c,d$ are integers.

Given,

$1000a+100b+10c+d=202(a+b+c+d)$

$\Rightarrow 798a-102b-192c-201d=0⟶\left(1\right)$

Also

$(1000b+100a+10d+c)+(1000a+100b+10c+d)=3333$

$\Rightarrow 1100b+1100a+11d+11c=3333$

$\Rightarrow 100b+100a+c+d=303⟶\left(2\right)$

Also, $a=c$ and $b=d$

$\therefore$ Equation $\left(1\right)$ reduces into,

$(798a-192a)-102b-201b=0$

$\Rightarrow 606a-303b=0$

$\Rightarrow 2a=b⟶\left(3\right)$

Also, equation $\left(3\right)$ reduces into

$(100a+a)+(100b+b)=303$

$\Rightarrow 101(a+b)=303$

$\Rightarrow a+b=3⟶\left(4\right)$

Combining equation $\left(3\right)$ and $\left(4\right)$ we get

$\therefore$ $3a=3$

$\Rightarrow a=1\Rightarrow c=1$

$b=2\Rightarrow d=2$

$\therefore$ the four digit number is $1212$.