Question

2) Without doing any calculation, find the numbers which are not perfect squares.
(i) 153
(ii) 257​

Answer 1

Both 153 and 257 are not perfect squares because the ones digits of these numbers are 3 and 7 respectively.

Theorem:- No perfect squares have the digits 2, 3, 7 and 8 in ones place.

Proof:- Given below is the congruency of various possible integers and their squares with theie remainders modulo 10.

$\displaystyle\longrightarrow\sf {10q\equiv0\implies (10q)^2\equiv 0\pmod{10}}$

$\displaystyle\longrightarrow\sf {10q+1\equiv1\implies (10q+1)^2\equiv 1\pmod{10}}$

$\displaystyle\longrightarrow\sf {10q+2\equiv2\implies (10q+2)^2\equiv 4\pmod{10}}$

$\displaystyle\longrightarrow\sf {10q+3\equiv3\implies (10q+3)^2\equiv 9\pmod{10}}$

$\displaystyle\longrightarrow\sf {10q+4\equiv4\implies (10q+4)^2\equiv 16\equiv6\pmod{10}}$

$\displaystyle\longrightarrow\sf {10q+5\equiv5\implies (10q+5)^2\equiv 25\equiv5\pmod{10}}$

$\displaystyle\longrightarrow\sf {10q+6\equiv6\implies (10q+6)^2\equiv 36\equiv6\pmod{10}}$

$\displaystyle\longrightarrow\sf {10q+7\equiv7\implies (10q+7)^2\equiv 49\equiv9\pmod{10}}$

$\displaystyle\longrightarrow\sf {10q+8\equiv8\implies (10q+8)^2\equiv 64\equiv4\pmod{10}}$

$\displaystyle\longrightarrow\sf {10q+9\equiv9\implies (10q+9)^2\equiv 81\equiv1\pmod{10}}$

Here none among 2, 3, 7 and 8 appears as the remainder, hence we can conclude that numbers ending in any of these four can't be a perfect square.

QED