Is the positive integer x divisible by each integer from 2 through 6 ?

The question asks whether x is divisible by the LCM of 2, 3, 4, 5, and 6, which is 60.

(1) x = 10m, where m is a positive integer divisible by each integer from 2 through 5.

m is divisible by the LCM of of 2, 3, 4, and 5, which is 60. Thus x = 10*(multiple of 60). Sufficient.

(2) 10x = n, where n is a positive integer divisible by each integer from 2 through 9.

n is divisible by the LCM of of 2, 3, 4, 5, 6, 7, 8, and 9, which is 2520 (9*7*8*5). Thus 10x = (multiple of 2520) --> x = (multiple of 252). If x = 252, then the answer is NO but if x is say, 252*60, then the answer is YES. Not sufficient.

Answer: A.
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considering the numbers one at a time in order to figure out what x will need, at a minimum, in it's prime factorization:

2: in order for x to be divisible by 2, x needs a 2 in its PF
3: in order for x to be divisible by 3, x needs a 3 in its PF (so now we have 2*3)
4: in order for x to be divisible by 4, x needs another 2 in its PF (so now we have 2*2*3)
5: in order for x to be divisible by 5, x needs a 5 in its PF (so now we have 2*2*3*5)
6: in order for x to be divisible by 6, x needs a 2 and a 3 in its PF (we already have that so our min doesn't change; x still needs, at min, 2*2*3*5 in its PF)

So we need enough info to prove that x has at least 2*2*3*5 in its PF

Statement 1.

x = (10)(m) ---> rewriting m in terms of what we know about its PF --> x=(2*5)(2*3*2*5*whatever else might be in m)

We can see right away that x has at least 2*2*3*5 in its PF. Sufficient.

Statement 2

10x = n ---> x = n/10 --> rewriting m in terms of what we know about its PF, and factoring 10 -->
x = (2*2*2*3*3*5*7*whatever else might be in n)/2*5 --> the two and 5 cancel to get --->
x = (2*2*3*3*7*whatever else might be in n)

We don't know if x has 2*2*3*5 in its PF because we don't know for sure that there is a 5; there may or may not be a 5 in the 'whatever else might be in n' part. Insufficient.

A is the answer.

Hello, can someone please explain to me why can we add another five in the second statement?
It's kind of driving me nuts!

Bunuel

I got this question in GMATPREP 5 exam. Please tag this to exampack 2

I have a simpler approach

m is div by 2 to 5
Write all primes then just see what's missing and add it into the list

SEE
m = 2x3x5 x(2) is the smallest div by all no.s 2-5

Similarly in the second one
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We need to check whether x is divisible by 4,5,6 (if yes, then definitely x is divisible by 2 and 3 as well)

St 1: x = 2 * 5 * m
Since "m is a positive integer divisible by each integer from 2 through 5"; then m is divisible by 6 as well. If m is divisible by 4,5, 6, then x is also divisible by 4,5,6.
Hence St 1 is sufficient by iself

St 2: x = n/10; we know that n is divisible by 4,5,6 but we cannot say for sure that x is also divisible by 5. Hence not sufficient.

Option A is the correct answer.
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