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

1. $1$
2. $2$
3. $3$
4. $4$

Let first number be $x$, and the second number be $y.$

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

The factors of $45$ are $15, 3, 9, 5, 1$ and $45$

Hence, the possible pairs of numbers are $(9,6), (7,2)$ and $(23,22).$

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

The correct answer is $(C).$
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$\textrm{3 pair,9 and 6 or 7 and 2 or 23 and 22. }$
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