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.
_________________

clearly A is suff....
1- x = 10m, where m is a positive integer divisible by each integer from 2 through 5
x=10m.. and m is a multiple of 2 to 5................
since m is a multiple of 2 and 3 both, it is also div by 6..
hence suff

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

if n = 2*3*4*5*6*7*8*9...
10x = 2*3*4*5*6*7*8*9...
so x = 3*4*6*7*8*9...
so x need not be div by 5...NO

if n = 2*3*4*5*6*7*8*9*5...
ans is YES
Insuff

A

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.

If we modify the original condition and the question, since x has to be divisible by 2, 3, 4, 5 and 6, the question becomes “is it x a multiple of 60?”
In case of con 1), from x=10m, m is divisible by 2, 3, 4 and 5. Hence, x is a multiple of 60. The answer is yes and the condition is sufficient. Thus, the correct answer is A.

- Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.
_________________

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

Hi,

St2: 10x = n --> x = n/10
n can be any integer.
If n = 2*3*4*5*6*7*8*9*5, then x is divisible by each integer from 2 through 6 --> Another 5 is added here to show sufficiency.
If n = 2*3*4*5*6*7*8*9, then x is not divisible by 5 because n/10 consumes a 2 and 5 --> In this case St2 is not sufficient
We can conclude that St2 is not sufficient at all times.

Hope it helps

Bunuel

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

_________________
Done. Thank you.
_________________

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
_________________

Excellent explanation brunel.

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.
_________________

I got a little confused with the wording of the question...

because 10m can be interpreted as (10)(m)=10m,
or it can mean 10m where m is a positive integer divisible by each integer from 2 through 5 -> 102, 103, 104, 105

No. 10m can only mean 10*m. If it were a three digit number, then it would be explicitly mentioned.
_________________

I did the same, but now I am asking myself.

How do I know that the LCM of the numbers from 2 to 9 is exactly the product of them?

How can we be sure that the LCM is smaller?

I'd also like to have the opinion of Bunuel if possible.

Bunuel, how could I know that 2 and 6 are included?
_________________