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.
Hope it helps.
I have no idea why this is a 95 % difficulty level question.
Just know that-
Each term in the expression (x+y)^n is divisible by x except for the last term which is y^n333^222 / 7
Try and bring the dividend to a form ( multiple of divisor +_ something). And usually such Qs are formed like that.
(333)^222 = (336-3)^222
Each term in the above expression is divisible by 7 ( since 336 is divisible by 7) but the last term which is 3^222
So now essentially our task is to find the remainder when 3^222 is divided by 7
Again the same routine
3^222 = (3^3)^74 = 27^12
or, (28-1)^12
All terms are divisible but the last i.e (-1)^12
12 being a positive power this is equal to 1
Answer E
Am I missing something ?
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