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1 vote
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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. $4$
in Quantitative Aptitude 9.3k points 31 523 813
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2 Answers

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$\textrm{3 pair,9 and 6 or 7 and 2 or 23 and 22. }$
3.5k points 4 10 63
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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 $(C).$
9.3k points 31 523 813
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