1 vote

How many pair of natural numbers are there, the differences of whose squares is $45$ ?

- $1$
- $2$
- $3$
- $4$

0 votes

Let first number is $’x’$, and second number is $’y’.$

$(x^{2} - y^{2}) = 45$

$\implies (x - y)(x + y) = 45$

Thus, the factors of $45$ possibles are $15, 3, 9, 5, 1$ and $45$

Hence, numbers are $9$ and $6$ or $7$ and $2$ or $23$ and $22$

$\therefore$ The number of such pairs $= 3.$

So, the correct answer is $(C).$

$(x^{2} - y^{2}) = 45$

$\implies (x - y)(x + y) = 45$

Thus, the factors of $45$ possibles are $15, 3, 9, 5, 1$ and $45$

Hence, numbers are $9$ and $6$ or $7$ and $2$ or $23$ and $22$

$\therefore$ The number of such pairs $= 3.$

So, the correct answer is $(C).$