Bunuel wrote:

Financier wrote:

What is the remainder when (18^22)^10 is divided by 7 ?

А 1

B 2

C 3

D 4

E 5

I think this question is beyond the GMAT scope. It can be solved with Fermat's little theorem,

which is not tested on GMAT. Or another way:

\((18^{22})^{10}=18^{220}=(14+4)^{220}\) now if we expand this all terms but the last one will have 14 as multiple and thus will be divisible by 7. The last term will be \(4^{220}\). So we should find the remainder when \(4^{220}\) is divided by 7.

\(4^{220}=2^{440}\).

2^1 divided by 7 yields remainder of 2;

2^2 divided by 7 yields remainder of 4;

2^3 divided by 7 yields remainder of 1;

2^4 divided by 7 yields remainder of 2;

2^5 divided by 7 yields remainder of 4;

2^6 divided by 7 yields remainder of 1;

...

So the remainder repeats the pattern of 3: 2-4-1. So the remainder of \(2^{440}\) divided by 7 would be the same as \(2^2\) divided by 7 (440=146*

3+

2). \(2^2\) divided by 7 yields remainder of 4.

Answer: D.

Just remember that

all terms in the expression (a+b)^n are divisible by a except for the last term i.e b^n. My solution uses just this much.

(18^22)^10 = (3*3*2)^22^10 = ((7-1)* 3 )^22^10 = [(7-1)^22 * 3^22]^10

(7-1)^10 divided by 7 will have Remainder = 1 --->

ANow for 3^22 =9^11= (7+2)^11 , the whole term will be divisible by 7 except 2^11

2^11= 2*2* 8^3

2*2* (7+1)^3 or, Remainder =4 ---->

BFrom statement A and B ,

18^22^10 = (Remainder 1*Remainder4)^10

4^10 = 4 * 64^3 = 4* (63+1)^3

4* remainder 1

Answer D

I need someone to validate this approach or did I just go absolutely berserk

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