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13 Feb 2025, 01:30
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Difficulty:
5%
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Question Stats:
93% (00:47) correct
7%
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wrong
based on 85
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If s is a positive integer greater than 0 and divisible by both 9 and 24, what is the least possible value of s?
(A) 36
(B) 48
(C) 72
(D) 96
(E) 144
(A) 36
(B) 48
(C) 72
(D) 96
(E) 144
13 Feb 2025, 02:19
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Approach 1:
LCM of 9 and 24 will be divisible by both
9 = 3^2
24 = 2^3*3
LCM = 3^2 * 2^3 = 72
Approach 2:
Also, can back-solve by checking the option which is divisible by both starting from top (since we need the smallest value)
(A) 36 - not divisible by 24
(B) 48 - not divisible by 9
(C) 72 - divisible by both
LCM of 9 and 24 will be divisible by both
9 = 3^2
24 = 2^3*3
LCM = 3^2 * 2^3 = 72
Approach 2:
Also, can back-solve by checking the option which is divisible by both starting from top (since we need the smallest value)
(A) 36 - not divisible by 24
(B) 48 - not divisible by 9
(C) 72 - divisible by both
Bunuel
If s is a positive integer greater than 0 and divisible by both 9 and 24, what is the least possible value of s?
(A) 36
(B) 48
(C) 72
(D) 96
(E) 144
(A) 36
(B) 48
(C) 72
(D) 96
(E) 144
13 Feb 2025, 22:29
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If s is a positive integer greater than 0 and divisible by both 9 and 24, what is the least possible value of s?
S will be divisible by all factors of 9 and 24
9 = 3*3
24= 3*8
Taking LCM, s should be devided by 3*3*8 = 72
Hence least no 72
IMO Ans C
S will be divisible by all factors of 9 and 24
9 = 3*3
24= 3*8
Taking LCM, s should be devided by 3*3*8 = 72
Hence least no 72
IMO Ans C
