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Divisibility:Gmatprep

by zaarathelab » Sun Dec 18, 2011 7:49 am
When positive integer n is divided by 3, the remainder is 2; when positive integer t is divided by 5, the remainder is 3. What is the remainder when the product is nt is divided by 15?

a) n-2 is divisible by 5

b) t is divisible by 3

I tried to do it algebraically, but I took a lot of time. What is the quickest way to do this (under 2 minutes)?

Are there any tips to know when to follow the algebraic method and when to follow the plugging in number method for divisibility/remainder problems?
Last edited by zaarathelab on Sun Dec 18, 2011 11:18 pm, edited 3 times in total.
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by GmatMathPro » Sun Dec 18, 2011 8:13 am
zaarathelab wrote:When positive integer n is divided by 3, the remainder is 2; when positive integer t is divided by 5, the remainder is 2. What is the remainder when the product is nt is divided by 15?

a) n-2 is divisible by 3

b) t is divisible by 3

I tried to do it algebraically, but I took a lot of time. What is the quickest way to do this (under 2 minutes)?
Start by listing possible values for n and t by starting with the remainders and adding the divisor repeatedly:

n = 2,5,8,11,14,...

t = 2,7,12,17,22,27...

Now, note that statement 1 is giving you absolutely no new information. We already know that n is two more than a multiple of 3, so clearly n-2 IS a multiple of 3. So we can eliminate A, C, and D and focus on statement 2.

Statement 2 says that t is divisible by 3, so let's restrict our attention to t=12,27...

Try some values for nt:

n=2, t=12 nt=24, remainder when divided by 15 is 9
n=5, t=12 nt=60, remainder when divided by 15 is 0

INSUFFICIENT.

Therefore, ans: E
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by zaarathelab » Sun Dec 18, 2011 9:47 am
Thanks Pete. That was beautiful.

My way was very lengthy

n= 3q+2 - - -(1)
t = 5p + 3 - - -(2)

Now, nt = (3q+2)(5p+3) = 15qp + 9q + 10p + 6 - - - - -(3)

Statement 1:

n-2 = 5k
Substituting this value in (1)

5k+2 = 3q+2
5k = 3q - - - - - -this proves that q is a multiple of 5. (4)

This info makes 9q divisible by 15 (15qp is already divisible), but gives us no info on 6p

INSUFFICIENT

Statement 2:

t=3s - -(5)
substituting this value in 2,

3s = 5p+3
p = 3(s-1) / 5 - - - - - this proves that p is a multiple of 3, making 10p a multiple of 3, hence 10p is divisible by 15.

But this gives us no info on 9q.

INSUFFICIENT.

Combining the two, remainder will always be 6
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by GmatMathPro » Sun Dec 18, 2011 9:56 am
Hmm, I just noticed that I think you made a couple typos when you put in the original problem. You wrote "a) n-2 is divisible by 3" but in your work it looks like you were using "n-2 is divisible by 5", and in fact, when i googled the problem i see that the second one is the correct one. Also, the original problem said t has a remainder of 3 when divided by 5, not 2. So, back to the drawing board...
Last edited by GmatMathPro on Sun Dec 18, 2011 10:44 am, edited 1 time in total.
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by GmatMathPro » Sun Dec 18, 2011 10:38 am
zaarathelab wrote:When positive integer n is divided by 3, the remainder is 2; when positive integer t is divided by 5, the remainder is 3. What is the remainder when the product is nt is divided by 15?

a) n-2 is divisible by 5

b) t is divisible by 3

I tried to do it algebraically, but I took a lot of time. What is the quickest way to do this (under 2 minutes)?

Are there any tips to know when to follow the algebraic method and when to follow the plugging in number method for divisibility/remainder problems?


Okay, so I think your solution is fine. Generally, if you're trying to prove sufficiency, an algebraic approach has some benefits, because you can never really PROVE sufficiency by showing that it works out for a few cases. However, we may be able to combine the approaches to make the purely algebraic approach less tedious.

From the given: n=3a+2, t=5b+2: n=2,5,8,11,14,17... t=3,8,13,18,23,28...

Statement 1: n-2 is divisible by 5, so: n=5c+2, n=2,7,12,17,... Combining this with the list for n above, we can see that n=15d+2. Once you find the first number that satisfies both conditions, each successive number will be LCM(5,3)=15 away, hence n=15d+2. Possible values of nt: n=2, t=3, nt=6 (remainder of 6); n=2, t=8, nt=16 (remainder of 1). INSUFFICIENT.

Statement 2: t is divisible by 3, so t=3f: t=3,6,9,12,... Combining with the list above, 3 is the smallest value that satisfies both conditions, so t=15g+3, so t=3,18,33... Possible values of nt: n=2,t=3 nt=6 (remainder of 6); n=5,t=3 nt=15 (remainder of zero) INSUFFICIENT.

Statements 1&2: n=15d+2 and t=15g+3. We can list numbers here if we want, but again, that can only make a strong case for sufficiency but it can't prove sufficiency, so it's risky. Algebraically, n*t=(15d+2)(15g+3) = 15^2*dg + 3*15d + 2*15d + 6. The first three terms are clearly divisible by 15, so the remainder of n*t when divided by 15 must always be 6. SUFFICIENT.

Whether this approach is preferable to a purely algebraic approach will depend on how comfortable you are manipulating the equations and interpreting the results quickly and accurately. If you're not that comfortable, listing some numbers can make you more confident that your equations make sense.
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by [email protected] » Sat Feb 18, 2012 5:42 am
Make this sum as simple as possible...

From the stimulus and statement 1 we arrive to the equation as: n = 15p + 2
now for example put as many values as possible 2, 17, 32 etc etc...


From the stimulus and statement 2 we arrive to the equation as: t = 15q + 3
now for example put as many values as possible 3, 18, 33, 48 etc etc...


Combined statements:

nt = (15p + 2) (15q + 3)

= 225pq + 45p + 30q + 6

As you can see that in the equation, the first terms are divisible by 15 but the last term is not

put as many values as possible, you will always get 6 as the remainder...

for eg 6, 306, 1056 etc etc...


the remainder will always be 6...

Hence the correct answer is C...


Hope this really helped...
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