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What is the remainder if 7^10 is divided by 100?

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What is the remainder if 7^10 is divided by 100?  [#permalink]

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New post 26 Oct 2010, 18:00
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Question Stats:

67% (01:10) correct 33% (01:02) wrong based on 372 sessions

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What is the remainder if 7^10 is divided by 100?

A] 1
B] 43
C] 19
D] 70
E] 49



I need a help from those who generally employ binomial thm way to attack these kind of problems, because I feel this is the shortest.
I generally try to bring numerator in a form that denm can divide it using binomial theorem i.e.
Num: (7^2)^5 =49^5 = (50-1)^5
Had the denm been 50, I would have easily got the remainder as ((-1)^5)/50 which is -1 or 50 - 1= 49 (Remainder)

But here, since the denm is 2*50, how should I employ this method?
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Re: What is the remainder if 7^10 is divided by 100?  [#permalink]

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New post 26 Oct 2010, 21:05
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the above expression can be written as (7^5)^2/10^2 making the power of the base and the numerator same now when you devide 7^5 by 10 the remainder is 7 as the unit digit is 7 for 7^5, so the expression becomes 7^2 = 49.
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Re: What is the remainder if 7^10 is divided by 100?  [#permalink]

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New post 27 Oct 2010, 00:00
1
ptm30 wrote:
What is the remainder if 7^10 is divided by 100?
A] 1
B] 43
C] 19
D] 70
E] 49



I need a help from those who generally employ binomial thm way to attack these kind of problems, because I feel this is the shortest.
I generally try to bring numerator in a form that denm can divide it using binomial theorem i.e.
Num: (7^2)^5 =49^5 = (50-1)^5
Had the denm been 50, I would have easily got the remainder as ((-1)^5)/50 which is -1 or 50 - 1= 49 (Remainder)

But here, since the denm is 2*50, how should I employ this method?


You have done 99% of the work to get the answer, here is the final bit :

You have established that 7^10 = 50k + 49

Bow there is two possibilities, either k is even or k is odd
If k is even, 7^10 = 100m + 49 (where k=2m) ...Hence remainder is 49
If k is odd, 7^10 = 100m +50 +49 (where k=2m+1) .. Hence remainder is 99

So remainder can only be 49 or 99, but since 99 isnt a choice, by POE, the answer must be 49

Answer : e
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Re: What is the remainder if 7^10 is divided by 100?  [#permalink]

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New post 27 Oct 2010, 01:58
Hi,

I understand what you say,
What if the answer options also had 99? In one of the very lengthy solutions that I read, 49 was the answer posted.
But my point is that I want to answer this question with certainty in the shortest possible way.

My search for the shortest solution is still on...
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Re: What is the remainder if 7^10 is divided by 100?  [#permalink]

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New post 27 Oct 2010, 02:10
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ptm30 wrote:
Hi,

I understand what you say,
What if the answer options also had 99? In one of the very lengthy solutions that I read, 49 was the answer posted.
But my point is that I want to answer this question with certainty in the shortest possible way.

My search for the shortest solution is still on...


You could always just solve this without the binomial theorem :

Last two digits of 7^2 = 49
Last two digits of 7^4 = 01 (7^2 * 7^2)
Last two digits of 7^5 = 07 (7^4 * 7)
Last two digits of 7^10 = 49 (7^5 * 7^5)

This will always be faster than the binomial approach

Note that the last two digits of a multiplication of two numbers only depend on the last two digits of those two numbers. So I don't need to calculate the full numbers
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Re: What is the remainder if 7^10 is divided by 100?  [#permalink]

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New post 23 Sep 2016, 21:04
7^2^5/100 =49^5/100

remainder is 49 Option E
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Re: What is the remainder if 7^10 is divided by 100?  [#permalink]

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New post 01 Oct 2017, 08:45
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ptm30 wrote:
What is the remainder if 7^10 is divided by 100?

A] 1
B] 43
C] 19
D] 70
E] 49



\(7^1 = 7\)
\(7^2 = 49\)
\(7^3 = 343\)
\(7^4 = 2401\)

\(\frac{7^4}{100} = Remainder \ 1\)

\(7^{10}\) = \(7^{4*2}*7^2\)

\(\frac{7^8}{100} = Remainder \ 1\)
\(\frac{7^2}{100} = Remainder \ 49\)

Hence, the remainder will be (E) 49
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Re: What is the remainder if 7^10 is divided by 100?  [#permalink]

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New post 28 Jan 2018, 07:42
shrouded1 wrote:
ptm30 wrote:
What is the remainder if 7^10 is divided by 100?
A] 1
B] 43
C] 19
D] 70
E] 49



I need a help from those who generally employ binomial thm way to attack these kind of problems, because I feel this is the shortest.
I generally try to bring numerator in a form that denm can divide it using binomial theorem i.e.
Num: (7^2)^5 =49^5 = (50-1)^5
Had the denm been 50, I would have easily got the remainder as ((-1)^5)/50 which is -1 or 50 - 1= 49 (Remainder)

But here, since the denm is 2*50, how should I employ this method?


You have done 99% of the work to get the answer, here is the final bit :

You have established that 7^10 = 50k + 49

Bow there is two possibilities, either k is even or k is odd
If k is even, 7^10 = 100m + 49 (where k=2m) ...Hence remainder is 49
If k is odd, 7^10 = 100m +50 +49 (where k=2m+1) .. Hence remainder is 99

So remainder can only be 49 or 99, but since 99 isnt a choice, by POE, the answer must be 49

Answer : e


shrouded1 Can you please tell me where I can find the explanation for this "binomial approach"? Thanks a lot :-)
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Re: What is the remainder if 7^10 is divided by 100?  [#permalink]

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New post 08 Jul 2018, 16:32
Another way to get to the answer:
7^10=100q+r
7^10=(7^5)^2. As I now know that 7^4=XX01;
=> Remainder = 1x7^2 =49
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Re: What is the remainder if 7^10 is divided by 100? &nbs [#permalink] 08 Jul 2018, 16:32
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