Question

Lcm of two natural numbers is 590 and their hcf is 59. how many sets of values are possible?

Answer 1

Solution :-

Since, H.C.F. of the two natural numbers = 59, so we can write -

A = 59x and B = 59y

Using H.C.F × L.C.M. = Product of two numbers

59x × 59y = 59 × 590

⇒ xy = 10

So, sets of possible numbers are = (1, 10,)(10, 1) and (2, 5)(5, 2)

Hence, 4 numbers sets of possible numbers are (59, 590), (590, 59), (118, 295) and (295, 118)

Answer 2

Given:

H.C.F. of the two natural numbers(A,B) = 59

L. C.M of two natural Numbers= 590

1st natural number (A )= 59x
& 2nd natural number( B) = 59y

H.C.F × L.C.M. = Product of two numbers

59 × 590 = 59x × 59y

(59×59)xy= 59 × 590

xy = 59× 590/ 59× 59

xy = 10

sets of possible numbers are = (1, 10,)(10, 1) and (2, 5)(5, 2)

Hence, 4 numbers sets of possible numbers are (59, 590), (590, 59), (118, 295) & (295, 118)


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Hope this will help you....