What is the hundreds digit of 201^201

Apart from cyclicity, we can read this as remainder when divided by 1000..
so \(201^{201}=(200+1)^{201}=200^{201}+.....200^2*1^{199}+200^1*1^200+1^{201}\)
so all terms apart from last two terms will be divisible by 1000..
hence remainder will be \(200^1*1+1=201\)
the hundreds digit therefore will be 2

C
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pattern of hundreds digit of \((201)^n\):-
2-4-6-8-0
Cyclicity:- 5

We have \(\frac{201}{5}\) leaves 1 as remainder.
So, hundreds digit of \((201)^{201}\) is same as hundreds digit of \((201)^1\).

Therefore, the required hundreds digit is 2.

Ans. (C)
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\((201)^0\) - Hundred's digit is 0
\((201)^1\) - Hundred's digit is 2
\((201)^2\) - Hundred's digit is 4
\((201)^3\) - Hundred's digit is 6
\((201)^4\) - Hundred's digit is 8
\((201)^5\) - Hundred's digit is 0
\((201)^6\) - Hundred's digit is 2
\((201)^7\) - Hundred's digit is 4

A pattern emerges and 8 is the hundred's digit for the 4,9,14,19.......199th power of 201.

Therefore, the hundred's digit of \(201^{201}\) is 2(Option C)

[quote="chetan2u"]What is the hundreds digit of \(201^{201}\) ?
(A) 0
(B) 1
(C) 2
(D) 5
(E) 7

Hundreds digit is 2.please correct if I am wrong



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PKN, how do you know the pattern is 2-4-6-8-0?

I solved this using binomial expansion.
If you think of 201^201 = (200+1)^201 = 201C0*200^201 + 201C1*200^200 + ... + 201C200*200^1 + 201C201*200^0
^the highlighted term equal 201*200 = 40200
^the highlighted part is the hundreds digit. All other terms will have higher powers and won't contribute to the hundreds term.

Answer: C

Hi sandman13,
If you observe carefully, We have each of the digit of 201 , lies between 0 and 2 inclusive. On successive powers of 201 from 1 to infinity, the tens digit of 201 decides the hundreds digit(1*any number=that number.no carry forward)). Here the tens digit is zero, hence 0*any number=0 with no carry forward.

Hundreds digit of \(201^1\)=2
Hundreds digit of \(201^2\)= 2+Hundreds digit of \(201^1=2+2=4\)
Hundreds digit of \(201^3\)= 2+Hundreds digit of \(201^2=2+4=6\)
Hundreds digit of \(201^4\)= 2+Hundreds digit of \(201^3=2+6=8\)
Hundreds digit of \(201^5\)= 2+Hundreds digit of \(201^4=2+8=0\)
Hundreds digit of \(201^6\)= 2+Hundreds digit of \(201^5=2+0=2\)
Hundreds digit of \(201^7\)= 2+Hundreds digit of \(201^6=2+2=4\)

This was my approach, it mayn't be a standard one.
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pattern of hundreds digit of (201)^(201):-
2-4-6-8-0
Cyclicity:- 5
So 201/5 remainder =1 so the hundred digit will be 2

Option C

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Is the OA really correct? Everybody seems to chose and provide explanations for C...

Incase C is incorrect, could someone provide an explanation as to why C is wrong?

Regards,
Chris
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The answer is (c)2.

My approach might be lengthy yet easy! Here it goes.

Note: 401, 601 etc are all the last 3 digits

\(201^{201}\) can be written as \(201^{200}\) * 201.
= \((201^{2})^{100}\) * 201
= \((401)^{100}\) * 201
= \((401^{2})^{50}\) * 201
= \(801^{50}\) * 201
= \(((801)^{2})^{25}\) * 201
= \((601)^{25}\) *201
= \((601^{2})^{12}\) * 201 * 601
= \((201)^{12}\) *201 * 601
= \(((201)^{2})^{6}\) *201 *601

again \(201^2\) is 401 hence substitute,
\(401^2\) is 801
= 601 * 801 * 201 * 601
the hundreds digit of the above is 2.

Thanks,
Uma

Hi Arro44,
OA is (C).
Do you have any points?

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