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]
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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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