If m is an integer, is m/102 an integer?
(1) 99 is a factor of 165m.
(2) 34 is a factor of 7m
[spoiler]OA=C[/spoiler]
Source: Veritas Prep
If m is an integer, is m/102 an integer?
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Prime factorize: 102 = 2*51 = 2*3*17
So we can be certain m/102 is an integer if we can be certain m is divisible by 2, 3 and 17.
Statement 1 tells us 165m is divisible by 99, or that 165m/99 is an integer. So (33)(5m)/(33)(3) is an integer, and 5m/3 is an integer. That means the '3' in the denominator cancels somehow with something in the numerator, and since it can't cancel with the '5', it must cancel with something in 'm'. So m must be divisible by 3.
Similarly Statement 2 tells us 7m is divisible by 34 = (2)(17), which can only be true if m is divisible by 2 and 17.
So the two statements together ensure m is divisible by 2, 3 and 17 and the answer is C.
So we can be certain m/102 is an integer if we can be certain m is divisible by 2, 3 and 17.
Statement 1 tells us 165m is divisible by 99, or that 165m/99 is an integer. So (33)(5m)/(33)(3) is an integer, and 5m/3 is an integer. That means the '3' in the denominator cancels somehow with something in the numerator, and since it can't cancel with the '5', it must cancel with something in 'm'. So m must be divisible by 3.
Similarly Statement 2 tells us 7m is divisible by 34 = (2)(17), which can only be true if m is divisible by 2 and 17.
So the two statements together ensure m is divisible by 2, 3 and 17 and the answer is C.
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