santorasantu wrote:
Is x divisible by 15?
(1) When x is divided by 10, the result is an integer
(2) x^2 is a multiple of 30
I have a following doubt, everybody is going for C for the answer and I think it is E.
the approach is as follows:
from 1: when x is divided by 10, the result is an integer. e.g. x= 20, 30, 40.....any number that ends in 0. In case of 30, x is divisible by 30 but when x = 20, then not,2 answers so NSF
from 2: x^2 is a multiple of 30. there for x^2 can be 60 or 90 or 120 or 150....which implies that x is sqrt(60) or sqrt(90) or sqrt (120)...when x is sqrt(60), then not divisible by 15, but when x = sqrt(900) = 30 ,then divisible by. therefore NSF.
combining 1 + 2: suppose x^2 is 60000 which implies x is sqrt(60000) = 100*sqrt(6). This number is divisible by 10 as per statement 1 and x^2 is a multiple of 30 as per statement 2. but not divisible by 15.
suppose x^2 is 900, then we have an answer that x is divisible by 15.
so 1+2 is also NSF.
I go for E, please let me know if I missed something
\(\sqrt{60,000}=100\sqrt{6}\) is NOT divisible by 10.
1. Every GMAT divisibility question will tell you in advance that any unknowns represent positive integers (ALL GMAT divisibility questions are limited to positive integers only).
2. On the GMAT when we are told that \(a\) is divisible by \(b\) (or which is the same: "\(a\) is multiple of \(b\)", or "\(b\) is a factor of \(a\)"), we can say that:
(i) \(a\) is an integer;
(ii) \(b\) is an integer;
(iii) \(\frac{a}{b}=integer\).
So, \(\sqrt{60,000}=100\sqrt{6}\) is NOT divisible by 10, because \(\sqrt{60,000}=100\sqrt{6}\) is NOT an integer and the result is also NOT an integer.
Is \(x\) divisible by 15?(1) When \(x\) is divided by 10, the result is an integer --> \(\frac{x}{10}=integer\) --> \(x=10*integer\). Now, if \(x=0\) (in case \(integer=0\)), then the answer is YES but if \(x=10\) (in case \(integer=1\)), then the answer is NO. Not sufficient.
From this statement though we can deduce that \(x\) is an integer (since \(x=10*integer=integer\)).
(2) \(x^2\) is a multiple of 30 --> if \(x=0\), then the answer is YES but if \(x=\sqrt{30}\), then the answer is NO. Not sufficient.
(1)+(2) Since from (1) \(x=integer\) then in order \(x^2\) to be divisible by 30=2*3*5, \(x\) must be divisible by 30 (\(x\) must be a multiple of 2, 3 and 5, else how can this primes appear in \(x^2\)?), hence \(x\) is divisible by 15 too. Sufficient.
Notice that \(x\) can be positive, negative or even zero, but in any case it'll be divisible by 30.Answer: C.
We edited this question and in the new GMAT Club tests this question reads:If x is a positive integer, is x divisible by 15?(1) x is a multiple of 10 --> if \(x=10\), then the answer is NO but if \(x=30\), then the answer is YES. Not sufficient
(2) x^2 is a multiple of 12 --> since \(x\) is an integer, then \(x^2\) is a perfect square. The least perfect square which is a multiple of 12 is 36. Hence, the least value of \(x\) is 6 and in this case the answer is NO, but if for example \(x=12*15\) then the answer is YES. Not sufficient.
Notice that from this statement we can deduce that \(x\) must be a multiple of 3 (else how can this prime appear in \(x^2\)?).
(1)+(2) \(x\) is a multiple of both 10 and 3, hence it's a multiple of 30, so \(x\) must be divisible by 15. Sufficient.
Answer: C.
Hope it's clear.