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1. How many pairs of natural numbers is there the difference of whose squares are 45.
Or, (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
So, number of such pairs = 3
Solution:
(x2 - y2) = 45Or, (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
So, number of such pairs = 3