If n is a positive integer, what is the remainder when n^10

Question

If n is a positive integer, what is the remainder when n^10 is divided by 12? 

(1) when n is divided by 12, the remainder is 2. 
(2) n is a multiple of 13.

ps_dahiya · Answer

A

St1: Let n = 12a + 2 where a is a +ve integer and 2 is remainder
So n^10 = (12a+2)^10. All the terms of this will be multiple of 12 except last one which will be 2^10 = 1024. rrmainder is 4: SUFF

St2: Nothing clear.: INSUFF

sgrover · Answer

(E) for me.

(A) can be ruled out easily.
try n = 2 and 14, with n = 2, remainder = 4
n = 14, remainder = 6
(B) is also insufficient. Try n = 13, 26

Combining,
n = 26, remainder = 6
n = 182, remainder = 4

alimad · Answer

If n is a positive integer, what is the remainder when n^10 is divided by 12? 

(1) when n is divided by 12, the remainder is 2. 
(2) n is a multiple of 13.


(1). n/12 = x*12 + 2 where x = quotient

If we apply simple logic n could be 14, 26, 38 so not suff

(2). n could be 13, 26,39,42,55 ..... not sufficient

Combining the two - 26, a viable candidate, multiple of 13 and when divided by 12 gives us remainder of 2. 

C my answer. Love to see a rebuttal.

acfuture · Answer

(1) n = 12*k + 2, since we have a remainder of 2.
so (12*k + 2)^10/12 =  (12*k)^10/12  +  2^10/12 . In red we see that it is multiple of 12 which has not remainders. In blue, 2^10 = 1024, when divided by 12, get remainder of 4. Suff...

(2) n=13,26,39,42.. remainder will be different values. Insuff...

A is my answer....

v1rok · Answer

Agree with (A) as the answer, except that

 (12*k + 2)^10/12 DOES NOT EQUAL (12*k)^10/12 + 2^10/12

For simple case, (a+b)^2=a^2+2ab+b^2. But as in this simple example,
the first two members of decomposition are divisible by a, so the remainder must come from the b^2. One can generalize this to ^10 case -- same logic.

ps_dahiya · Answer

How did you reach at the conclusion that (14^10)/12 will give you a remainder of 6????

I am sure answer should be A.
As I said, if n/12 gives you a remainder of 2 then n can be represented as 
n = 12a+2
and n^10 = (12a+2)^10
If you solve this then all terms will have atleast one 12a and one term will be without 12a and that will be 2^10.
So the remainder will be equal to remainder of 2^10/12 i.e 1024/12 i.e 4.

For example (a+b)^3 = a^3+b^3+3ab^2+3a^2b. All terms are divisible by "a" except b^3.

sgrover · Answer

oops.. sorry for the confusion.. Somehow I have been reading it as "what is the remainder when n^10 is divided by  10 instead of 12 "..

Sorry again for the confusion.
A is the anwer.

jaynayak · Answer

A.

1) n = 12Q + 2
Hence n^10 will have remainder 2. Infact n^x will have remainder 2 when x>0

2) n is a multiple of 3 

N can be 26 in that case reminder is 2 

N can be 39 in that case remiander is 3

Not suff

acfuture · Answer

arrrhhhh... thanks for the correction...

If n is a positive integer, what is the remainder when n^10

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