N is the product of all integers from 1 to 75, inclusive. If 10^k is

Question

N is the product of all integers from 1 to 75, inclusive. If 10^k is a factor of N, what is the greatest possible integer value of K?

A. 5
B. 10
C. 12
D. 18
E. 25

tam87 · Accepted Answer

N = 75!

we need to find the highest power of 10 in 75!
as 10 = 2 x 5, we need to find highest power of 5 only, as 5s will be always less than 2s.

highest power of 5 = 75/5 + 75/25 = 15 + 3 = 18 (option D)

JeffTargetTestPrep · Answer

We see that N = 75! and know that k is an integer.

75!/10^k = 75!/ ^k = 75!/  = integer

The expression above is an integer if k is not greater than the total number of 2s in the prime factored form of 75! and if k is not greater than the total number of 5s in the prime factored form of 75!

Since, in the prime factored form of 75!, there are fewer 5s than 2s, we need to concentrate on the total number of 5s, which we can quickly determine with the following technique:

75/5 = 15

15/5 = 3

The total number of 5s is 15 + 3 = 18, so it must be true that:

k ≤ 18

We see that the maximum value of integer k is 18.

Answer: D

Kinshook · Answer

Given: N is the product of all integers from 1 to 75, inclusive. 
Asked: If 10^k is a factor of N, what is the greatest possible integer value of K?

N=75!
Highest power of 5 in 75! = 15 + 3 = 18
Highest power of 2 in 75! = 37 + 18 + 9 + 4 + 2 + 1 = 71

Highest power of 10 in 75! = min (18,71) = 18

IMO D

Khushi2102 · Answer

Why are we taking 10^k as 5^k X 2^k and not 10^k itself. The answer could be 7

Posted from my mobile device

Paras96 · Answer

Successively dividing 75 by 2

we get 75/2= 37 /2= 16 /2= 9 /2= 4 /2= 2 /2= 1

Power of 2 in 75! = 37+18+9+4+2+1=71

Successively dividing 75 by 5

we get 75/5= 15 /5= 3

Power of 5 in 75! = 15+3=18

Therefore power of 10 in 75! = minimum (Power of 2 in 75!,Power of 5 in 75!)=minimum(71,18)=18

Hence D

Markkawin · Answer

Ok so this question is very easy with the formula (n/p)+ n/(p^2)+n/(p^3)+… but i struggle to find where it came from so this is my explanation for the logic behind this formula:
1. Given that n! = 1x2x…x(n-1)x(n)
2. If you want to find whether an integer i is the factor of n you can:
2.1 find the prime multiplier of the number i (all the little building blocks of i)
2.2 find the prime multiplier of the number n (all the little building blocks of n)
3. if prime multipliers of n contains all the prime multipliers of i then i is a factor of n 
4. But don’t forget that n! Is the product of every numbers less than n so this means that i might also be a factor of these numbers! So how to find what are these numbers without writing them all out? Well, since we know i then the first number is i itself, the 2nd is 2i and so on until just before or equal to n. Thus when we write n/p, the quotient is the number of times i is the factor of n!. And surely if n very large, i^2 might be a factor of n! as well. Thus we find out the number of times i^2 is a factor of n! by finding the quotient of n/(p^2) in the formula. Repeat with p^3,p^4,… until quotient is <1 (denominator is no longer factor of n)
****BUT there is a catch here this only works when i is prime number!!!!! Any composite number should be broken down into product of prime numbers and the greatest prime number should be used. This is because there will be least number of the greatest prime number compared to other prime in n! and since this is n! Every prime building blocks of numbers are multiplied together and so we CANNOT treat composite number as one entity. These primes building blocks are all connected my multiplications regardless its original number******
5. When you add all the quotients up then this gives you the power of i that is the highest factor of n!

- So back to the question, what is the factor we want to find is 10^k this is broken down into (5^k)(2^k) and n! is 75!
- how many 5 are in 75! -> 75/5 = 15 therefore there are 15 5s in 75!
- how many 5^2 are in 75! -> 75/25 = 3 therefore there are 3 5^2 in 75! (Half of the 5 in 5^2 was counted as part of 15 5s)
- 5^3>75 there is no 5^3 in 75!
> therefore there are total of 15+3 = 18 5s in 75! -> k=18

Posted from my mobile device

youcouldsailaway · Answer

I found the best way to tackle this problem to be as follows:

N/2^k + N/2^k +N/2^k

We will take only quotient values from above and add all those.

Now, we are asked to find how many times 10^k appears in 75!. We should look how many times prime factors of 10 appear in the factorial as it is basically the same as finding how many times 10 appears in the factorial. Plus, while 2 and 5 may appear elsewhere (with other numbers like 4 and 15) we can still group 2 and 5 together to make a 10, which we would have overlooked had we only focused on how many times 10 appears in the factorial.

Let's look at 2 first:

75/2 = 37
75/4 = 18
75/8 = 9
75/16 = 4
75/32 = 2
75/64 = 1

Therefore 2 appears 71 times. Note that the denominators are all different powers of 2.

Let's look at 5 now:

75/5 = 15
75/25 = 3
75/125 = 0

Therefore 5 appears 18 times. Note that the denominators are all different powers of 5.

Now, we know that we can only make a 10 when we have both 5 and 2. So, to make a 10 we only have 18 5s and so we will need 18 2s.

Thus, answer is D.

Credit to Shubham from Quora.

danru07 · Answer

Could someone explain why we have to consider 5^2?

N is the product of all integers from 1 to 75, inclusive. If 10^k is

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N is the product of all integers from 1 to 75, inclusive. If 10^k is

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