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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, 04:26
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A
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D
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Question Stats:

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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, 11:22.
Last edited by EgmatQuantExpert on 20 Dec 2018, 04:26, 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 20 Dec 2018, 00:20
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

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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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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, 22:52

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 ....  [#permalink]

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New post 04 May 2019, 04:16
EgmatQuantExpert, chetan2u

I see that your solution is quite complex.
I just want to offer some alternative.

1) What digits are recurring in 7^x? I will mark the last digit as [x]
7^1 = 7
7^2 = 4[9]
7^3 = 4[9]*7 = 9*7 = [3]
7^4 = [3]*7 = [1]
The same as in 7^1
Hence pattern repeats every 4 products (the last digit of 7^5 is the same as 7^1)
As a consequence, 7^16 has [1] as a last digit

So, when we deal with 100!, we deal with something*100, which divides by 7^16.
Since 7^16 has 1 as a last digit and 100! has [00] as a last digits we see that division is full. No remainders.
When we deal with 100!+100 the remainder will be 100.
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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 04 May 2019, 04:39
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Arseniy8 wrote:
EgmatQuantExpert, chetan2u

I see that your solution is quite complex.
I just want to offer some alternative.

1) What digits are recurring in 7^x? I will mark the last digit as [x]
7^1 = 7
7^2 = 4[9]
7^3 = 4[9]*7 = 9*7 = [3]
7^4 = [3]*7 = [1]
The same as in 7^1
Hence pattern repeats every 4 products (the last digit of 7^5 is the same as 7^1)
As a consequence, 7^16 has [1] as a last digit

So, when we deal with 100!, we deal with something*100, which divides by 7^16.
Since 7^16 has 1 as a last digit and 100! has [00] as a last digits we see that division is full. No remainders.
When we deal with 100!+100 the remainder will be 100.



That may not be the correct way to do it..
You have to follow [109/7]+[100/7^2]....,
If it were 7^20, you will again get 1 as units digit, and 100! Would be divisible by 7^20, as per your method.
But that is not correct, as 100! is not divisible by 7^20 as there are only 16 sevens in 1*2*3*..*99*100.
So 100! Is divisible by 7^1, 7^2...till 7^16 but not by 7^17 or 7^20 and higher powers
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Re: If N is the product of all integers from 1 to 100, both inclusive ....   [#permalink] 04 May 2019, 04:39
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