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Which of the following must be an integer if the integer x is divisible by both 18 and 21?

A. x/252
B. x/189
C. x/126
D. x/108
E. x/81

the integer has to be multiple of LCM of 18 and 21, which is 126..
so x/126 will be an integer
ans C

252 is also a multiple of both 18 and 21

I've seen a similar question being solved with primes but I am not sure how exacly:
18= 2 * 3^2
21= 3*7

Is it 2*3^2 *7?!
Can somebody explain this to me? Thanks

Hi,
the Q asks us 'must' be an integer....
we know x is div by 18 and 21, so to answer 'must' question, we take the worst scenario that x is the lowest common multiple..

say if x is 126, as per the conditions given..
will x/252 be an integer..NO
it will be 126/252=1/2.. so x/252 will not be always an integer, But the question asks us 'must' be an integer so it has to be always an int..

Hope it helps
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Bunuel
Which of the following must be an integer if the integer x is divisible by both 18 and 21?

A. x/252
B. x/189
C. x/126
D. x/108
E. x/81


x is divisible by both 18 and 21
189 is a factor of 18*21
==> B is an answer
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Bunuel
Which of the following must be an integer if the integer x is divisible by both 18 and 21?

A. x/252
B. x/189
C. x/126
D. x/108
E. x/81


x is divisible by both 18 and 21
189 is a factor of 18*21
==> B is an answer

18=2*3^2
21=3*7
LCM of 18 and 21 = 2*3^2 * 7 = 126

Hi nguyenphamphuong
x/189 -- if x here is 189 , then it won't be divisble by 18 in the first place.
18*21= 378 = 126*3 = LCM *3
LCM 126 is not divisible by 189 .

x/126 will always be an integer
Answer C
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