Bunuel wrote:

jonyg wrote:

what is the reminder when 333^222 is divided by 7?

a.3

b.2

c.5

d.7

e.1

official answer=>e

source- random internet

What is the remainder when 333^222 is divided by 7?A. 3

B. 2

C. 5

D. 7

E. 1

333^{222}=(329+4)^{222}=(7*47+4)^{222}. Now if we expand this, all terms but the last one will have 7*47 as a multiple and thus will be divisible by 7. The last term will be

4^{222}=2^{444}. So we should find the remainder when

2^{444} is divided by 7.

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;

...

The remainder repeats in blocks of three: {2-4-1}. So, the remainder of

2^{444} divided by 7 would be the same as

2^3 divided by 7 (444 is a multiple of 3).

2^3 divided by 7 yields remainder of 1.

Answer: E.

Hey can you explain to me how you get a remainder of 2 when you divide 2^1/7?