100! is divisible by 160^n?

Question

100! is divisible by 160^n...what is the max. integral value of n?

Ans : 19

what is the logic to solve it ?

Answer 1

First factor 160 into its prime factors: 160 = 2^5 * 5. So, the only numbers we need to consider are 2 and 5.

Now, in 100!, how many factors of 2 are there? Let's see... 50 for each even number, an extra 25 for each multiple of 4, an extra 12 for each multiple of 8, an extra 6 for each multiple of 16, 3 for each multiple of 32, and 1 for 64. Hence there are 50 + 25 + 12 + 6 + 3 + 1 = 97 factors of 2 in 100!, i..e. 160! = 2^97 * . Since each multiple of 160 require five 2s, that means 97/5 = 19 (the remainder has to be dropped, clearly) is the largest power of 2^5 that divides 100!

Checking the 5s works the same way, except the count is easier to get: 20 for each multiple of 5, 4 for each multple of 25, and that's it -- 24 in all. 100! = 5^24 * .

So, we see the 2s establish the limit for n, and n = 19.

Answer 2

First, factoring 160 we get 160 = 2^5 * 5. If we factor 100! , we get 100! = 2^p * 5^q * m, where p>=0 and q>=0 are integers and m is a positive integer that does not contain the prime factors 2 and 5. So, in order for 100! to be divisible by 160^n, we must have

5n <=p and n <= q. Therefore, we have to find p and q.

factor 2 - To find how many integers from 1 to 100 have the factor 2 with exponent1, we multiply 2 by each odd number from 1 to 49; to find how many have 2 with exponent 2, we multiply 4 by each odd number from 1 to 25, and so on. We form the following table:

1 2 25
2 4 13
3 8 6
4 16 3
5 32 2
6 64 1

That is, 25 numbers have the factor 2 with exponent 1, 13 with exponent 2.......3 with exponent 5. So, p = 25*1 + 13*2 + 6*3 + 3*4 + 2*5 + 6*1 = 97. This implies that 5n <= 97 => n<= 19. So, 19 is one of our bound for n.

Now, we have to count how may times the prime factor 5 appears in the factorization of 100! The numbers that have 5 with exponent 1 are obtained multiplying 5 by the integers that are not multiples of 5, that is, 1,2,3,4,6,7,9,11,12.....etc, until we get 100. The numbers that have 5 with exponent 2 are obtained multiplying 5^2 = 25 by these same numbers provided the product is <=110. And so on. So, we form the following table:

1 5 16
2 25 4

So, q = 16*1 + 4*2 = 24.

And we have n <= q = 24, so 24 is our second bound for n. It follows the maximum value for n is maximum{19, 24} = 19.

Answer 3

160 = 16*10 = 2^4*2*5 = 2^5*5

Therefore if we want 160^n, there need to be enough factors in 100! that can give us 2^(5n)*5^n.

In what follows, all of my division ignores remainders, which is called "integer division" in computer programming circles - I don't know about mathematical terms for that.

It's easier to count how many 5s are available. 100/5 = 20, so that's 20 fives right there, one for each mutliple of 5, such as 5, 10, 15, etc. 100/25 = 4, so that gives us 4 more or 24 total since we have to count an extra 5 for each number that is a multiple of 25. And 100/125 will give us no extra 5s.

So as far as the fives are concerned, we could do 160^24 and be alright. So we need to find out how many factors of 2 we have.

This should be equal to 100/2 + 100/4 + 100/8 + 100/16 + 100/32 + 100/64 = 50 + 25 + 12 + 6 + 3 + 1 = 97

Since 160 has five factors of 2, the highest n we can raise it to and still be a factor of 100! will be 97/5 = 19. If we tried to raise 160^20, we would have 100 factors of 5 while 100! only has 97.

--charlie

Answer 4

you should count how many 2 's and 5 's you have in 100!
as the answer is 19 i think you have 95 two's and 19 five's
you know 160 = (2^5) * 5 (^ sign means to the power of)

I exactly calculatet how many 2 `s and 5`s is in 100!
it is exactly as this:we have 97 , 2`s and 24, 5`s
we should find out how many fives is there for two`s
so : 97 / 5 = 19

Answer 5

Without doing any math here I am willing to be that 19 is the answer because anything above 19 would be larger than half of 100!....just a thought.

Answer 6

correct..!