DS - Numbers

agemroy · Posted: Tue Jul 08, 2008 10:05 am Post subject: DS - Numbers

If k, m, and t are positive integers and k/6 + m/4 = t/12 , do t and 12 have a common factor greater than 1 ?
(1) k is a multiple of 3.
(2) m is a multiple of 3.

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.

OA is A

AleksandrM · Posted: Tue Jul 08, 2008 10:46 am Post subject:

Simplify and you have: 2K + 3M = T

1) K is at least 3 so 2(3) + 3(1) = T = 9
2(3) + 3(2) = T = 12 and so on and so forth, whatever M is, T and 12 will always have a common factor greater than 1, because any number that is multiplied by three will have 3 as its factor (duh) and when it is added to a number that has a factor of 3, then the two will have a common factor of 3. Sufficient.

2) M is at least 3 so 2(1) + 3(3) = 11 No common factor
2(1) + 3(6) = 20 A common factor of 2. Not sufficient.

Choose A.

zakir · Posted: Tue Jul 08, 2008 10:29 pm Post subject:

I didn't get this Sad

how come you're plugging numbers for m (1,2,etc..) and k?

AleksandrM · Posted: Wed Jul 09, 2008 5:15 pm Post subject:

lunarpower · Posted: Thu Jul 10, 2008 11:37 pm Post subject: Re: DS - Numbers

yeah, so you should definitely rephrase the problem by getting rid of the fractions first. you can kill them all by multiplying through by 12, the common denominator of all the fractions:
2k + 3m = t
this is the base equation from which we'll be working to answer the problems.

you can also rephrase the question. if t and 12 have a common factor greater than 12, that's just another way of saying that something that goes into 12 (besides 1) also goes into t. so, here's a rephrase of the prompt question:
does 2, 3, 4, 6, or 12 go into t?
since 4, 6, and 12 are just compounds of the prime factors 2 and 3, it's actually redundant to ask whether those numbers are factors of t. therefore, we can rephrase the question down to its most distilled essence:
does 2 or 3 go into t?

-- statement (1) --
if k is a multiple of 3, then we can write k = 3n, where n is an integer. therefore, we have
2(3n) + 3m = t
6n + 3m = t
since 6n and 3m are both multiples of 3, it follows that 3 goes into t.
sufficient.

-- statement (2) --
if m is a multiple of 3, then we can write m = 3n, where n is an integer. therefore, we have
2k + 3(3n) = t
2k + 9n = t
these (2k and 9n) don't have a common factor, so this is insufficient.
if you want proof of insufficiency, go ahead and plug some easy numbers: if k = n = 1 (which would actually mean k = 1 and m = 3), then t = 11, which doesn't share a common factor with 12. if k = 3 and n = 1 (which would actually mean k = m = 3), then t = 15, which shares the common factor 3 with 12.
insufficient.

answer = a