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Bunuel
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GRE 1: Q169 V154
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Bunuel
If x is divisible by 2, 7, and 11, which of the following must also be a divisor of x?

A. 308
B. 154
C. 70
D. 44
E. 21

Since 2,7 & 11 have nothing in common, LCM of 2,7 & 11 must divide x

LCM of 2,7 & 11 is 154

Hence, Answer will be (B) 154
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In this both A and B looks similar.
308-2,2,7,11
154-2,7,11
but the question is about which other number can also be a divisor, in that case choosing B(154) seems wise as X has another divisor which is not that number itself.
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Bunuel
If x is divisible by 2, 7, and 11, which of the following must also be a divisor of x?

A. 308
B. 154
C. 70
D. 44
E. 21


If x is divisible by 2, 7, and 11, then x is either the LCM of 2, 7, and 11 or a multiple of the LCM. The LCM of 2, 7, and 11 is 2 x 7 x 11 = 154. Thus, x must be either 154 or a multiple of 154. In other words, 154 must be a divisor of x.

Answer: B
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x is divisible by 2, 7, and 11 so x must be divisible by The least commun multiple of 2, 7 and 11

Hence it is 154

Answer: B
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