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What is the remainder when 7^100 is divided by 50?

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What is the remainder when 7^100 is divided by 50?  [#permalink]

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New post 24 May 2018, 18:29
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[GMAT math practice question]

What is the remainder when \(7^{100}\) is divided by \(50\)?

\(A. 0\)
\(B. 1\)
\(C. 7\)
\(D. 21\)
\(E. 49\)

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Re: What is the remainder when 7^100 is divided by 50?  [#permalink]

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New post 24 May 2018, 19:03
MathRevolution wrote:
[GMAT math practice question]

What is the remainder when \(7^{100}\) is divided by \(50\)?

\(A. 0\)
\(B. 1\)
\(C. 7\)
\(D. 21\)
\(E. 49\)


This question essentially asking what are the last 2 digits of the expression \(7^{100}\)... as what ever is in the hunderds digit, if the last two are 00, the number is always divided by 50.
Now cyclicity of 7 is 4, And the numbers are 7,9,3,1........ Hence the last digit is 1 as 25*4=100
now \(7^{4}\) = 7*7*7*7= 2401
And the expression is \(2401^{25}\)= in that case the last two will be always 01 ( can be tested quickly with \(101^{2}\) & \(101^{3}\))
Hence the reminder is 01.....................Hence , I would go for option B.
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Re: What is the remainder when 7^100 is divided by 50?  [#permalink]

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New post 24 May 2018, 19:04
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Option B

\(7^1\) = 7
\(7^2\) = 49
\(7^3\) = 343
\(7^4\) = 2401
\(7^5\) = 16807
\(7^6\) = 117649
\(7^7\) = 823543
\(7^8\) = 5764801

...

So, 7^100 is something that ends with 01.

The remander is 1.
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Re: What is the remainder when 7^100 is divided by 50?  [#permalink]

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New post 24 May 2018, 20:35
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These type of questions become really simple if you understand the concept of negative remainders. Always try and reduce the dividend to 1 or -1.

= Rem [7^100 / 50]

= Rem [49^50/50]

= Rem [ (-1)^50 / 50]

= Rem [1 / 50]

= 1
hence B
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Re: What is the remainder when 7^100 is divided by 50?  [#permalink]

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New post 27 May 2018, 18:21
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The remainder when \(7^{100}\) is divided by \(50\) depends only on the units and tens digits.

The units digits of \(7^n\) cycle through the four values \(7, 9, 3\), and \(1\).
The tens digits of \(7^n\) cycle through the four values \(0, 4, 4\), and \(0\).

We have the following sequence of units and tens digits for \(7^n\):

\(7^1 = 07 ~ 07\)
\(7^2 = 49 ~ 49\)
\(7^3 = 343 ~ 43\)
\(7^4 = 2401 ~ 01\)
\(7^5 = 16807~ 07\)


So, \(7^{100} = (7^4)^{25}\) has the same units and tens digits as \(7^4\), that is, \(01\).
Thus, the remainder when \(7^{100}\) is divided by \(50\) is \(1\).

Therefore, B is the answer.

Answer : B
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Re: What is the remainder when 7^100 is divided by 50?  [#permalink]

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New post 29 May 2018, 09:24
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MathRevolution wrote:
[GMAT math practice question]

What is the remainder when \(7^{100}\) is divided by \(50\)?

\(A. 0\)
\(B. 1\)
\(C. 7\)
\(D. 21\)
\(E. 49\)


We see that 7^2 = 49, which is 50 - 1. Although 49/50 = 0 R 49, rather than using the remainder of 49, let’s call the remainder “-1”.

Since 7^100 = (7^2)^50 = 49^50, which is equivalent to (-1)^50 when it’s divided by 50, and since (-1)^50 = 1, so when (-1)^50 is divided by 50, the remainder is 1.

Answer: B
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What is the remainder when 7^100 is divided by 50?  [#permalink]

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New post 27 Aug 2018, 06:43
\(7^{100}/50=49^{50}/50=(50-1)^{50}/50\) - only \((-1)^{50} = 1^{50}=1\) - won't be devisable by 50. The remainder is 1.

Answer B.
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Re: What is the remainder when 7^100 is divided by 50?  [#permalink]

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New post 23 Sep 2018, 05:00
ScottTargetTestPrep wrote:
MathRevolution wrote:
[GMAT math practice question]

What is the remainder when \(7^{100}\) is divided by \(50\)?

\(A. 0\)
\(B. 1\)
\(C. 7\)
\(D. 21\)
\(E. 49\)


We see that 7^2 = 49, which is 50 - 1. Although 49/50 = 0 R 49, rather than using the remainder of 49, let’s call the remainder “-1”.

Since 7^100 = (7^2)^50 = 49^50, which is equivalent to (-1)^50 when it’s divided by 50, and since (-1)^50 = 1, so when (-1)^50 is divided by 50, the remainder is 1.

Answer: B



Hi,

thanks for this solution, but I have a doubt. this question doesn't say that there is exponent for 50. then, How can we take (-1)^50 ?

Waiting for reply.

Regards,
Kishlay
GMAT Club Bot
Re: What is the remainder when 7^100 is divided by 50? &nbs [#permalink] 23 Sep 2018, 05:00
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