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

Question

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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EgmatQuantExpert · Accepted Answer

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! =    +  = 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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chetan2u · Answer

N is divisible by all numbers till 100, so let us check how many 7s are there...
 + =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

Arseniy8 · Answer

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  
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.

chetan2u · Answer

That may not be the correct way to do it..
You have to follow  + ....,
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

Onyedelmagnifico · Answer

100! = 2^97 × 3^48 × 5^24 ×7^16 
 × N
Where N is product of powers of prime numbers above 7 and below 100.
Hence 100! Divides 7^16 
Since 100<7^16 hence 100 is a remainder.
Thus D.

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CrackverbalGMAT · Answer

This is a question where one has to know about the Maximum power concept and remainder concepts.

N is the product of all integers from 1 to 100, bot inclusive, means that N is 100!

Therefore, N + 100 can be written as 100! + 100. When this is divided by  7^{16} , we will have to find the remainder.

We can split the dividend into two parts and divide them individually by  7^{16}  and add the respective remainders to obtain the final remainder.

When 100! is divided by  7^{16} , the remainder is 0. This is because the highest power of 7 in 100! is 16 and hence the dividend is completely divisible by  7^{16} . 
When 100 is divided by  7^{16} , the remainder will be 100 itself, since 100 is smaller than  7^{16} .

Therefore, final remainder = 0 + 100 = 100. 
 The correct answer option is D .

Hope that helps!

Kinshook · Answer

Asked: 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} ?

N = 100!

Max power of 7 in 100! = 14 + 2 = 16
Therefore 100! is divisible by 7^{16}

Remainder when N= 100! is divided by 7^{16} = 0
Remainder when N + 100 is divided by 7^{16} = 100

IMO D

shubhim20 · Answer

please elaborate ...Why is this zero?

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

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