Question

If the sum of two numbers is 1215 and their H.C.F is 81,find at least 4 pairs of such numbers.

Answer 1

It is given that the sum of two numbers is 1215 and their H.C.F is 81

Since the H.C.F of two numbers is 81, the two numbers are 81x and 81y.

Now, 81x + 81y = 1215 [Sum of two numbers is 1215]

>> 81 (x + y) = 1215

>> (x + y) = 15

For x = 7, y = 8; in this case the numbers are 7 × 81 = 567 and 8 × 81 = 648

x = 1, y = 14; in this case the numbers are 1 × 81 = 81 and 14 × 81 = 1134

x = 2, y = 13; in this case the numbers are 2 × 81 = 162 and 13 × 81 = 1053

x = 4, y = 11; in this case the numbers are 4 × 81 = 324 and 11 × 81 = 891

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