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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.
Can someone please help me to solve this question??
Thank you!!!!
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.
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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(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
(1) Remainder of n/5 is 4 (2) Remainder of n/6 is 5
(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.
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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68% (02:01) correct 
