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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 .

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Thank you so much!!! Angels!!
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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

Answer C
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cjcj
What is the remainder of n/15?

(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.

Hence C answer.
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What is the remainder of n/15?

(1) Remainder of n/5 is 4
(2) Remainder of n/6 is 5

Fact 1 : N = 5*(Interger) + 4 So, N=4/9/14... Remainder of 4/15 = 4 is not equal to remainder of 9/15 = 9 Not Sufficient
Fact 2 : N = 6*(Interger) + 5 So, N =5/11/16... Remainder of 5/15 = 5 is not equal to reminder of 11/15 =11 Not Sufficient

Combining 1 & 2
5*(Integer) + 4 = 6*(Integer) +5
First common term = 29 second term = 29 + LCM(5,6) = 59.
Both yield the same remainder when divided by 15. 29/15 remainder = 14, 59/15 remainder = 14. sufficient
C
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