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The product of two numbers is 6760 and their H.C.F. is 13. How many such pairs can be formed?
3
4
2
1
Let the numbers be 13x and 13y. So, 13X ×13 Y = 6760 X×Y=40. Possible values of (x, y) are (1, 40); (2, 20); (4, 10); (5, 8) Only two acceptable values are (1, 40) and (5, 8).
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Anuj Kanojia
Why only two values acceptable ?
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