QUESTION

CSAT

Hard

Reasoning

Prelims 2020

How many pairs of natural numbers are there such that the difference of whose squares is 63?

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Explanation

Equation: (x2y2=63)(x^2 - y^2 = 63) Factoring: (xy)(x+y)=63(x - y)(x + y) = 63 Factor pairs of 63: (1,63)(1, 63), (3,21)(3, 21), (7,9)(7, 9) For each factor pair:

(xy=1,x+y=63)x=(64/2)=32,y=(6332)=31(x - y = 1, x + y = 63) \rightarrow x = (64 / 2) = 32, y = (63 - 32) = 31 \rightarrow Pair: (32,31)(32, 31) (xy=3,x+y=21)x=(24/2)=12,y=(2112)=9(x - y = 3, x + y = 21) \rightarrow x = (24 / 2) = 12, y = (21 - 12) = 9 \rightarrow Pair: (12,9)(12, 9) (xy=7,x+y=9)x=(16/2)=8,y=(98)=1(x - y = 7, x + y = 9) \rightarrow x = (16 / 2) = 8, y = (9 - 8) = 1 \rightarrow Pair: (8,1)(8, 1)

Result: 3 pairs

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