The LCM of two positive integers A and B = 590 and their HCF = 59. How

Question

The LCM of two positive integers A and B = 590 and their HCF = 59. How many set of values of A and B are possible?

A. 0
B. 3
C. 4
D. 5
E. 9

chetan2u · Answer

HCF=59, so let A=59x and B=59y, where x and y are co-prime.

Now product of two numbers = product of their LCM and HCF.

59x*59y=590*59
x*y=10=1*10=10*1=2*5=5*2

Unordered pair of integers that fit in are (1,10), (2*5), (5,2) and (10,1).

C

armaankumar · Answer

Prime Factorization of 590= 59*2*5 
Now, let's see how many pairs of A and B are possible with LCM 590 and HCF 59:
(59, 2*5);(59*5, 2);(59*2,5 );(59*2*5, 59)

There are 4 such sets possible.

ScottTargetTestPrep  Is this the correct way to solve this problem

sanjitscorps18 · Answer

The main idea of this question is to recognize that A and B have individual identities. This confusion is somehow removed as 2 is not one of the answer choices.

Overall we have GCF as 59 and LCM as 590. Hence we need to have two numbers that already have 59 common in them.

Since 59 has to be both part of A and B, we can manipulate 2 and 5

Keeping 59 aside, the following pairs can be formed (1, 10), (10, 1), (2, 5), (5, 2)

Actual number pairs would be formed by multiplying 50 with each, hence (59, 590), (590, 59), (108, 295), (295, 108)

Option C

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The LCM of two positive integers A and B = 590 and their HCF = 59. How

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The LCM of two positive integers A and B = 590 and their HCF = 59. How

Prep Toolkit

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