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.

acfuture · Answer

given n>0

st 1.
n= 14, 26, 38, 50, 62, 74, 86.... and so on... Insuff.

st 2.
n=13, 26, 39, 52, 65, 78, 91... and so on.. Insuff..

Together, n = 26...
(26^10)/12 = remainder of 3

there is no way that I am going to calculate 26^10, so I tried to find a pattern instead... I am pretty sure that there is a formula for this...

(26/12) => R=3
(26^2)/12 => 676/12 => R=3

therefore,
C is my answer...

sgrover · Answer

(C)

I. n can be 14, 26 .. Insufficient
II. n can be 13, 26 .. Insufficient

Combining, only value that satisifies both is 26

now that we know the value of n (=26), we can calculate the remainder, we dont need to calculate..

acfuture · Answer

you are right... i spent way too much time then....

Zooroopa · Answer

I propose a different explanation - Ans is  A .

Take (I):

Since n(mod12) = 2 ==> n = 12k + 2 , where k is an integer.
Now n^10 = (12k + 2)^10. Expand this. There would be 11 terms in the expression.  Note that the first term is a multiple of 12 .  Therefore the first 10 terms would all contain some power of 12 . So all of them are divisible by 12. Now we only have 2^10 remaining. We can always find the remainder of  2^10 (mod 12) . So (I) holds good.

Take (II):

n is a mutliple of 13 and n^10 would have to be divided by 13. Since 12 and 13 are co-primes there wouldn't be any pattern in the remainders. So there cannot be a single remainder for all possible multiples of 13. So (II) does not hold good.

shahnandan · Answer

My logic was similar to that of zaroopa's ..I second A.

acfuture · Answer

I have a question... bear with me...
in this case is it true to say that, whatever value of n^10 = (12k + 2)^10 if divided by 12, since it leads to the same remainder all the time, A is sufficient? Therefore, I did not have to find a specific value of n right?

Zooroopa · Answer

Right. We don't need to find the exact value ever! We need to ensure that we can find   a value   (this is very important as multiple values are going to kill the option).

ps_dahiya · Answer

Good one. Never thought of that. Kevin, could you please post the OA and OE?

jaynayak · Answer

Will go with A.

n/12 has a remainder of 2. hence the remainder of n^10 can be calculated by adding the individual remainders

For cond 2)
Since n is a multiple of 13. N can have any value and each value will give out a diff remainder when divided by 12
henc INsuff

MA · Answer

suppose n = 2. when n^2 divided by 12, r results 4, which we can say that 2+2 = 4. correct. 
now lets divide n^3 = 8 by 12. r is 8 not 6. 

how do you apply your approach??? 

this approach seems faulty. it would be true if 10n is divided by 12.

sharadGmat · Answer

good one.. I just confirmed using calculator... 14^10, 26^10,38^10 all leave the same remainder of 4.. i.e. remainder when 2^10 is divided by 12.

kevincan · Answer

Zooroopa, wonderful work! As for (2), n may or may not be a muliple of 12. If it is, the remainder when n^10 is divided by 12 is 0. If it is not, it will NOT be 0. OA is A. What do people mean by OE? (I made this question up)

kevincan · Answer

I guess if I'm the author I should provide an OE, then. But what Zooroopa said will certainly do! OE sounds so final, though :-D

Zooroopa · Answer

Hey kevin - good question pal! Please post a few of these supposedly 'authentic ones' 8-)

!!

MA · Answer

i. n can be 14, 26, ...., 338, .... 494, ..... 808, ....
ii. n can be 13, 26, ...., 338, .... 494, ..... 808, ....

there are infinite numbers of values that satisfiy statement 1 and 2.

chiragr · Answer

I picked A--used with same logic as Zooroopa's logic.

I have seen variation of question but logic stays same Consider follwoing example.

lets say A and B when devided by 12 leaves remainder of R1 and R2
So, A=12x+r1
 B = 12y+r2
A*B = (12x+r1)*(12y+r2)
 = 12x*12y + 12x*r2 +12y*r1 + r1*r2
 = 12(12xy+xr2+yr1) + R
can be represented as = 12z + R

so if we know r1 and r2(R) we know remainder of A*B.

here we can say in A that 
n=12x+2 or y+2 where y=12x

n^10 = (y+2)^10 expansion of this will have terms which are power of y(y^2 or y^3) power of 2, or combination of some power of y and 2. 

so we can represent 12z + R where R is multiple of 3 and will be constant for any value of n when n^10 is expanded.

hence A sufficient

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

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