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31 Dec 2021, 10:30
Kudos
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
56% (01:33) correct
44%
(01:42)
wrong
based on 228
sessions
History
Date
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What is the least possible common multiple of two distinct integers greater than 120?
A. Product of 121 and 122
B. 500
C. 250
D. 242
E. 121
A. Product of 121 and 122
B. 500
C. 250
D. 242
E. 121
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10 Jan 2022, 04:15
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Bunuel
Thank you @Experts' Global for such new questions.
Elaborating a bit more on the video explanation already given.
Since we need the smallest LCM of two integers that are greater than \(120\)
First:
indentify the smallest integer that is greater than \(120\) in this case it is \(121\)
Second :
Prime factorize the smallest integer obtained above i.e. \(121 = 11 *11\)
Third :
Multiply the smallest positive integer possible, to the prime factorization obtained above to get smallest LCM i.e. \( = 11 *11 *2 = 242\)
In third step above the smallest positive integer CANNOT be \(1\) as mutiplying \(1\) will lead to the same integer.
We need two distinct integers hence first integer is \(121 ( 11 *11)\) and next integer is \(242 ( 11 *11 *2)\)
So in summary :
The two distinct integers are \(121\) and \(242\)
and the smallest LCM is \(242.\)
Ans D
Hope it's clear.
