Question
Solution
As HCF is 13 so let numbers be 13x & 13y. It means x & y must be co-prime. Now 13x × 13y = 10140, So xy = 60. Now pairs of (x, y) can be (1,60) , (2,30) , (3,20) , (4,15) , (5,12) & (6, 10). Here only (1,60) , (3,20) , (4,15) & (5, 12) are co-prime so number of pairs are 4.
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