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by ceilidh.erickson » Fri Sep 11, 2015 12:36 pm
Whenever you see "remainder," you know the question is testing properties of divisibility.

If n is a positive integer greater than 6, what is the remainder when n is divided by 6?

Target question: is n divisible by 6 (meaning divisible by 2 and 3)? And if not, do we know enough about n to determine the remainder?

(1) n^2 - 1 is not divisible by 3.

First, factor the difference of squares: n^2 - 1 --> (n + 1)(n - 1)
This tells us that the product of the number one greater than n (n + 1) and one less than n (n - 1) is NOT divisible by 3. That must mean that n itself IS divisible by 3. (If you take any 3 integers in a row, 1 of them will be a multiple of 3 and the other two won't).

So, we know that n is divisible by 3, but we don't know whether it's divisible by 2. Insufficient.

(2) n^2 - 1 is even.

We can factor this the same way: (n + 1)(n - 1) must be even, and therefore divisible by 2. Since n + 1 and n - 1 are 2 apart, this tells us that they must both be even, and therefore n must be odd.

This by itself is not enough to answer the question, since n could be 7, giving us a remainder of 1 when divided by 6, or 9, giving us a remainder of 3 when divided by 6. Insufficient.

1&2 together: n is odd and a multiple of 3. Therefore, the remainder must always be 3 when divided by 6, because an odd multiple of 3 is always 3 more than a multiple of 6 (an even multiple of 3).

The answer is C.
Last edited by ceilidh.erickson on Sat Sep 12, 2015 8:50 am, edited 1 time in total.
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by rommysingh » Fri Sep 11, 2015 1:40 pm
Ceilidh thanks for the reply but can you show by example like taking real time values in case 2
2) n^2 - 1 is even.

We can factor this the same way: (n + 1)(n - 1) must be even, and therefore divisible by 2. However, we don't know whether it's also divisible by 3. Maybe it's a multiple of 6, maybe not. Insufficient.

How?
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by MartyMurray » Fri Sep 11, 2015 8:43 pm
rommysingh wrote:Ceilidh thanks for the reply but can you show by example like taking real time values in case 2
2) n^2 - 1 is even.

We can factor this the same way: (n + 1)(n - 1) must be even, and therefore divisible by 2. However, we don't know whether it's also divisible by 3. Maybe it's a multiple of 6, maybe not. Insufficient.

How?
That does not look right actually. It looks as if Ceilidh jumped from thinking about n to thinking about n² - 1.

If (n + 1)(n - 1) is even, then either n + 1 or n - 1 has to be even. If one of them is even, then both of them are even and n is not even.

If n = 7, n - 1 = 6 and n + 1 = 8.

So what Statement 2 is actually telling us is that n is odd. Any odd integer greater than 6 will work. So we cannot from Statement 2 determine the remainder when n is divided by 6 because the remainder has multiple possible values. For example, if n = 7, r = 1. If n = 11, r = 5.

Combined the question and the statements tell us the following.

n > 6 (from the question)
n is a multiple of 3. (from Statement 1)
n is odd. (from Statement 2)

So n is an odd multiple of 3 that is greater than 6, for example, 9, 15, or 21. Therefore the reminder when n is divided by 6 will always be 3.

So combined the statements are sufficient and the answer is C.
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by [email protected] » Fri Sep 11, 2015 10:16 pm
Hi rommysingh,

This question can be solved by TESTing VALUES, although you'll have to do a little bit of 'brute force' math to define the patterns involved.

We're told that N is a POSITIVE INTEGER greater than 6. We're asked for the REMAINDER when N is divided by 6.

1) N^2 - 1 is NOT divisible by 3

IF...
N = 7, then 49 - 1 = 48, which IS divisible by 3 (so N CANNOT be 7)

IF...
N = 8, then 64 - 1 = 63, which IS divisible by 3 (so N CANNOT be 8)

IF...
N = 9, then 81 - 1 = 80, which is NOT divisible by 3. 9 divided by 6 is 1 remainder 3 (so the answer to the question is 3).

At this point, you should notice the potential pattern in the work that you've done so far (the first two integers that we TESTed did NOT fit what we were told, but the third integer did. Chances are likely that this pattern continues...)

IF...
N = 10, then 100 - 1 = 99, which IS divisible by 3 (so N CANNOT be 10)

IF...
N = 11, then 121 - 1 = 120, which IS divisible by 3 (so N CANNOT be 11)

IF...
N = 12, then 144 - 1 = 143, which is NOT divisible by 3. 12 divided by 6 is 2 remainder 0 (so the answer to the question is 0).
Fact 1 is INSUFFICIENT

2) N^2 - 1 is EVEN

From our prior work (in Fact 1 - above), we've already worked through several examples. Under what conditions will N^2 - 1 be EVEN...

IF...
N = 7, then 49 - 1 = 48, which is EVEN. 7 divided by 6 is 1 remainder 1 (so the answer to the question is 1).

IF...
N = 9, then 81 - 1 = 80, which is EVEN. 9 divided by 6 is 1 remainder 3 (so the answer to the question is 3).
Fact 2 is INSUFFICIENT

Combined, we know...
N^2 - 1 is NOT divisible by 3 (so every third value will "fit"
N^2 - 1 is EVEN

We already have one TEST that 'fits' both Facts...

IF...
N = 9, then 81 - 1 = 80. 9 divided by 6 is 1 remainder 3 (so the answer to the question is 3).

From here, we know that N = 12 is NOT going to 'fit' Fact 1, so we need to jump to the next 'third value'...

IF...
N = 15, then 225 - 1 = 224, which is EVEN and NOT divisible by 3. 15 divided by 6 is 2 remainder 3 (so the answer to the question is 3).

After doing all of this work, it's highly unlikely that the same remainder would be a coincidence. If you wanted to do one more TEST though, you could...

IF...
N = 21, then 441 - 1 = 440, which is EVEN and NOT divisible b 3. 21 divided by 6 is 3 remainder 3 (so the answer to the question is 3).

It certainly appears that the answer to the question is ALWAYS 3...
Combined, SUFFICIENT

Final Answer: C

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by ceilidh.erickson » Sat Sep 12, 2015 8:51 am
Marty - you're totally right! Even experts can make mistakes when they're moving too quickly. I've edited the post above.
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by Max@Math Revolution » Mon Sep 14, 2015 2:30 am
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem.
Remember equal number of variables and independent equations ensures a solution.


If n is a positive integer greater than 6, what is the remainder when n is divided by 6?

(1) n^2 - 1 is not divisible by 3.

(2) n^2 - 1 is even.

In the original condition there is 1 variable (n) and we need 1 equation to match the number of variable and equation. Since there is 1 each in 1) and 2), normally D would have a high chance of being the answer. And for rest of the question it would be better for us to directly substitute the values.

In case of 1), if n=9,12,...... and are multiples of 3, n=9=6*1+3, n=12=6*2+0 would gives us an answer that is not unique. therefore the condition is not sufficient.

In case of 2), n^2-1=even, n^2=even-1=odd, n=odd thus we have values n=7,9,11,... while n=7=6*1+1, n=9=6*1+3.. and therefore not unique. The condition is not sufficent. Using both 1) & 2) together, we have n=9,27,45... thus n=9=6*1+3, n=27=6*4+3, n=45=6*7+3. Since the remain is 3, the answer is unique and therefore the conditions are sufficient. Therefore the answer is C.

( Normally for cases where we need 1 more equation, such as original conditions with 1 variable, or 2 variables and 1 equation, or 3 variables and 2 equations, we have 1 equation each in both 1) and 2). Therefore D has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) separately. Here, there is 59 % chance that D is the answer, while A or B has 38% chance. There is 3% chance that C or E is the answer for the case. Since D is most likely to be the answer according to DS definition, we solve the question assuming D would be our answer hence using 1) and 2) separately. Obviously there may be cases where the answer is A, B, C or E.)

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