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If N is the product of all integers from 1 to 100, both inclusive ....

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If N is the product of all integers from 1 to 100, both inclusive ....  [#permalink]

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New post Updated on: 20 Dec 2018, 03:26
00:00
A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

57% (01:25) correct 43% (01:57) wrong based on 74 sessions

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If N is the product of all integers from 1 to 100, both inclusive, then what is the remainder when N + 100 is divided by \(7^{16}\)?

    A. 0
    B. 7
    C. 49
    D. 100
    E. 1343

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Originally posted by EgmatQuantExpert on 19 Dec 2018, 10:22.
Last edited by EgmatQuantExpert on 20 Dec 2018, 03:26, edited 1 time in total.
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Re: If N is the product of all integers from 1 to 100, both inclusive ....  [#permalink]

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New post 19 Dec 2018, 23:20
EgmatQuantExpert wrote:
If N is the product of all integers from 1 to 100, both inclusive, then what is the remainder when N + 100 is divided by \(7^{16}\)?

    A. 0
    B. 7
    C. 49
    D. 100
    E. 1343

To read all our articles:Must read articles to reach Q51

Image


N is divisible by all numbers till 100, so let us check how many 7s are there...
[100/7]+[100/7^2]=14+2=16, so N is divisible by 7^(16).
But what about N+100...
N is divisible, so let us check for 100..
7^(16) is >100, so 100 will be the remainder..

D
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
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3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: If N is the product of all integers from 1 to 100, both inclusive ....  [#permalink]

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New post 01 Jan 2019, 21:52

Solution


Given:
We are given that,
    • N = 1 * 2 * 3 * 4 * … * 99 * 100

To find:
We are asked to find out,
    • The remainder when N + 100 is divided by \(7^{16}\)

Approach and Working:
    • We know that,\(R{\frac{(N + 100)}{7^{16}}} = R(\frac{N}{7^{16}}) + R(\frac{100}{7^{16}})\)
      o R\((\frac{N}{7^{16}}) = R\frac{(1 * 2 * 3 * 4 * .... * 99 * 100)}{7^{16}} = R(\frac{100!}{7^{16}})\)

    • Power of 7 in 100! = \([\frac{100}{7}] +[\frac{100}{7^2}] = 14 + 2 = 16\)
      o So, \(100! = 7^{16} * k\), where k is an integer

    • Thus, \(R(\frac{100!}{7^{16}}) = 0\), and
    • \(R(\frac{100}{7^{16}}) = 100\)

Therefore, the remainder when N + 100 is divided by \(7^{16}\) is 0 + 100 = 100

Hence the correct answer is Option D.

Answer: D


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Re: If N is the product of all integers from 1 to 100, both inclusive .... &nbs [#permalink] 01 Jan 2019, 21:52
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