Question

How many $4$-digit numbers are there with the property that it is a square and the number obtained by increasing all its digits by $1$ is also a square?

• A. $0$
• B. $1$
• C. $2$
• D. $4$

Answer 1

Answer: (B)

$1$

Let original four digit number is = $m^2$
and new 4-digit number is = $n^2$
then $m^2-n^2=1111$ (where m, n, $\epsilon$ I)
$(m+n)(m-n)=1\times 1111$ or $11\times 101$
Case - I
$m+n=1111$
$m-n=1$
$m=556$
$n=555$
But $m^2=$ is a 4-digit number
Case - II
$m+n=101$
$m-n=11$
$m=56$ and $n=44$
So, there is only one such 4-digit number.