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e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 2950
If N is the product of all integers from 1 to 100, both inclusive ....  [#permalink]

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18 00:00

Difficulty:   65% (hard)

Question Stats: 57% (01:59) correct 43% (02:15) wrong based on 121 sessions

### HideShow timer Statistics 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

_________________

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.
Math Expert V
Joined: 02 Aug 2009
Posts: 7763
Re: If N is the product of all integers from 1 to 100, both inclusive ....  [#permalink]

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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

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).
N is divisible, so let us check for 100..
7^(16) is >100, so 100 will be the remainder..

D
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e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 2950
Re: If N is the product of all integers from 1 to 100, both inclusive ....  [#permalink]

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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.

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Intern  B
Joined: 13 Aug 2018
Posts: 12
Re: If N is the product of all integers from 1 to 100, both inclusive ....  [#permalink]

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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
7^3 = 4*7 = 9*7 = 
7^4 = *7 = 
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  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  as a last digits we see that division is full. No remainders.
When we deal with 100!+100 the remainder will be 100.
Math Expert V
Joined: 02 Aug 2009
Posts: 7763
Re: If N is the product of all integers from 1 to 100, both inclusive ....  [#permalink]

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1
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
7^3 = 4*7 = 9*7 = 
7^4 = *7 = 
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  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  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..
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
_________________ 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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