Two numbers both greater than 29, have HCF=29 and LCM=4147. The sum of the numbers is
A
666
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B
669
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C
696
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D
966
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Solution
The correct option is C696 Product of numbers =29×4147 Let the numbers be 29a and 29b.
Then, 29a×29b=24×4147 ⇒ab=143 Now, co-primes with product 143 are (1,143),(11,13) So the prime numbers are (29×1,29×143)and(29×11,29×13) Since both numbers are greater than 29, the suitable pair is (29×11,29×13) ⇒(319+377)=696.