Hi cjcj,

(1) First statement tells us that n=5q+4, where q is a non-negative integer. You can tell right away this alone is insufficient when dividing by 15.

(2) Second statement tells us that n=6p+5, where p is a non-negative integer. You can tell right away this alone is insufficient when dividing by 15.

Combining both statements, you can find a common equation. See Bunuel's post to see how to do that: when-positive-integer-n-is-divided-by-5-the-remainder-is-90442.html#p722552

First few terms of n, when n=5q+4, is 4,9,14,19,24,29,34
First few terms of n, when n=6p+5, is 5,11,17,23,29,35

First common term is 29. So the general equation will be a multiple of the lcm of 5 and 6, which is 30 plus 29. So the new equation of n after combining is n=30z+29, where z is a non-negative integer. At this point it should be obvious that this statement is sufficient because n is equal to 29 more than a multiple of 15.

n=15*(2z)+15+14 -->n=15*(2z+1)+14, so because n is equal to 14 more than a multiple of 15, the remainder will always be 14.

I actually don't even factor as I have done above. I like to look at the individual remainders of the terms when dividing by the divisor and then combine using the operation in question and then find the remainder after that. For example:
(30*z)/15 yields a remainder of 0. 29/15 yields a remainder of 14. Combine 0 and 14 by addition because the operation in question is addition and then divide by 15 and find the remainder.

For example let's say you wanted to find the remainder of \(\frac{(18*23)}{7}\). Mod(\(\frac{18}{7}\)) is 4. Mod(\(\frac{23}{7}\)) is 2. Combine 4 and 2 by multiplication because the operation in question is multiplication and then find the remainder. So the remainder of \(\frac{(4*2)}{7}\) is 1 and this is the answer.

This can be done with negative remainders too. For example let's say you wanted to find the remainder of \(\frac{(23-18)}{7}\). Combine 2 and 4 by subtraction and then find the remainder. So the remainder of \(\frac{(2-4)}{7}\) is 5. When you have a negative remainder, find the remainder first as if the numerator was positive and then subtract from the divisor.
If you had positive 2 on top, the remainder would have been 2, but you have negative 2, so subtract positive 2 from 7 and you get a remainder of 5.

Hi cjcj,
If we write down values that will agree to the statements gievn we can solve it easily

Statement 1:
n can be 9, 14, 19, 24, 34, 39, 44, 49, 54, 59...
Each of these numbers will give different remainders when divided by 15
So INSUFFICIENT

Statement 2:
n can be 5, 11, 17, 23, 29, 35, 41, 47, 53, 59...
Each of these numbers will give different remainders when divided by 15
So INSUFFICIENT

Combining both statements:
n could be 29, 59, 89, 119
We will get a common number in the list after every 30 numbers, as the LCM (5, 6) = 30
The Remainder of 29, 59, 89 etc. when divided by 15 will always be same.

SUFFICIENT

Statement 1:

st 1:
n/5 --remainder is 4 .... so list of remainder values can be .
4, 9, 14, 19, 24,29, 34, 39, 44, 49, 54, 59...
Each of these numbers will give different remainders when divided by 15
So INSUFFICIENT

Statement 2:
n/6 --remainder is 5 .....so list of remainder values can be 5, 11, 17, 23, 29, 35, 41, 47, 53, 59...
Each of these numbers will give different remainders when divided by 15
So INSUFFICIENT

Combining both statements:
check for the first common remainder ....which is 29 -> from st 1 and st 2

We will get a common number in the list after every 30 numbers, as the LCM (5, 6) = 30
The Remainder of 29, 59, 89 etc. when divided by 15 will always be same.

SUFFICIENT together ...hence C .

please give kudos if you like the post .
Regards,,
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Thank you so much!!! Angels!!

(1) n=5*k+4, where k>=1
n can take values = 9 14 19
remainders after division by 15 = 9 14 4
Insufficient
(2) n=6*k+5, where k>=1
n can take values = 11 17 23
remainders after division by 15 = 11 2 8
Insufficient
(1)+(2) Let's express both equations in terms of -ve remainders:
n=5*k-1
n=6*k-1
so n = 30*k-1
reexpressing it with positive remainder:
n=30*k+29
testing division by 15
n=(30*k+29)/15, we can conclude that 30*k will always be divisible by 15; 29 will leave 14 as a remainder, so here is the answer.
Sufficient

Answer C

(1) n = 5x+4 (where x is a non negative integer). But this doesnt tell us about 15. Insufficient.

(2) n = 6y+5 (where y is a non negative integer). But this also doesnt tell us about 15. Insufficient.

Combining the two statements, n = 5x+4 = 6y+5
This tells us that when n is divided by 5, remainder is 4 and when n is divided by 6, remainder is 5. We can notice that the difference between divisor and remainder in each case is constant as '1' (5-4 = 1 = 6-5)

Now there is a rule - If a number n gives remainder R1 when divided by D1, and a remainder R2 when divided by D2; such that D1-R1 = D2-R2 = a constant C,
then the number n can be written generally as:

n = (LCM of D1, D2)*k - C (where k is a positive integer)

So as per this rule, for our question, we can write n as:
n = (LCM of 5&6)*k - 1 = 30K - 1.

As we can see, 30K-1 means 1 less than multiple of 30, and since every multiple of 30 is also a multiple of 15, so 1 less than multiple of 30 is also 1 less than a multiple of 15. So we know that our given number is 1 less than a multiple of 15, hence we can find the remainder when divided by 15. Sufficient.

Hence C answer.

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