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Question

If 14 is the HCF of 4p and 21q, then minimum value of p and q is:



A

1

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B

2

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C

2, x

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D

No common factors exist

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Solution

The correct option is A

1


Given that,

14 is HCF of 4p and 21q.

Therefore, both of them should have 7 and 2 as prime factors.

4p=2×2×p

21q=3×7×q

So, p=7,4p=4×7=28, which is a multiple of 14.

Also, q=2,21q=21×2=42, which is a multiple of 14.

Therefore, p should be 7 and q should be 2.


Mathematics

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