Math, asked by Palak16716, 13 hours ago

The LCM and the HCF of two numbers are 1001 and 7 respectively. How many such pairs are possible?

(a)0
(b)1
(c)2
(d)7

[EXPLAIN IN EASY WAY & BRIEFLY]​

Answers

Answered by tanu20054
1

hope this will help you...

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Answered by anjumanyasmin
2

Given:

LCM=1001

HCF=7

Let the two numbers be x and y.

According to the question we have

x=7a and y=7b, (where a and b are co-primes)

x × y = LCM × HCF

7a × 7b = 1001 × 7

7ab = 1001

ab=1001/7

ab=143

(a , b)= (1,143) or (11,13)

The correct option is "c"

Hence there are two pairs are possible

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