N! is having 37 zeros at its end. How many values of N is/are possible?A: 0B: 1C: 5D: InfiniteThe correct answer is c.Can you explain this answer?

Question

Answer

Easiest way to answer this is,

Here it's given N! Ends with 37 zeros.
It becomes 38 zeros when the N cross a number which is divisible with 5 or 10
If N is 84. It isn't. But just assume.
If 84! Has 37 zeros,
83,82,81,80 factorial will have 37 zeros.
If it reach 85, it will become 38 zeros.
So N has five values.
We can generalize it,
Take any number of zeros 1 or 25 or 66.
There exist 5 digits whose factorial ends with those many zeros.

Answer

//First check the number of zeroes in factorials of 100. //
• So,
100/5 = 20(quotient)
now divide that quotient again by 5
that is,
• 20/5 = 4(quotient)

• now add both the quotient you get 24, which is the number of zeroes for 100!

• then 37 zeroes must be higher than 100!

• lets take a number which is easily divisible by 5.
• that is say, 150!
• now,
with the same formula,
• 150/5= 30(quotient)

• 30/5 = 6

• 6/5 = 1.2
• now lets add all these and see if we find 37 zeroes..
• 30 + 6 + 1.2 ~ 37 (0.2 is negligible)

• now to check the limit how far it can go until it reaches 38.. we will take an assumption of next number which is greater than 150 and divisible by 5.

• if you try finding the zeroes in 155! you will get 38 zeroes.

• 155/5 = 31

• 31/5 = 6.2

• 6/5 = 1.2 31+ 6+ 1= 38

{{{{
• 154/5= 30.8

• 30.8 /5 = 6.16

• 6.16/5= 1.03
• which doesn't reach 38 zeroes so 154! is safe.
}}}}

• so therefore only factorials 150, 151, 152, 153, 154 have 37 zeroes.

Answer

Can anyone please elaborate the solution

Answer

C

Answer

Solution:

To find the number of values of N, we need to find the range of N.

Let's consider the prime factorization of N.

The number of zeros at the end of N! is determined by the number of factors of 5 in the prime factorization of N!.

Therefore, we need to find the number of values of N such that the number of factors of 5 in the prime factorization of N! is equal to 37.

Let's consider the powers of 5 in the prime factorization of N!.

The number of factors of 5 in the prime factorization of N! is given by the formula:

n = floor(N/5) + floor(N/25) + floor(N/125) + ...

where floor(x) is the largest integer less than or equal to x.

Since we need to find the number of values of N such that n = 37, we can use the formula to find the range of N.

Let's start by finding the lower bound of N.

If n = 37, then we have:

37 = floor(N/5) + floor(N/25) + floor(N/125) + ...

Since each term in the sum is an integer, we can use the inequality:

floor(N/5) + floor(N/25) + floor(N/125) + ... ≤ N/5 + N/25 + N/125 + ...

Substituting into the original equation, we get:

37 ≤ N/5 + N/25 + N/125 + ...

37 ≤ N(1/5 + 1/25 + 1/125 + ...)

37 ≤ N(1/4)

148 ≤ N

Therefore, the lower bound of N is 148.

Let's find the upper bound of N.

If n = 37, then we have:

37 = floor(N/5) + floor(N/25) + floor(N/125) + ...

Since each term in the sum is non-negative, we can use the inequality:

floor(N/5) + floor(N/25) + floor(N/125) + ... ≥ floor(N/5)

Substituting into the original equation, we get:

37 ≥ floor(N/5)

Therefore, the upper bound of N is 185.

Therefore, the range of N is 148 ≤ N ≤ 185.

Let's count the number of values of N in this range.

We can count the number of values of N by counting the number of multiples of 5, 25, 125, etc. in the range.

There are 37 multiples of 5, 7 multiples of 25, and 1 multiple of 125 in the range.

Therefore, the total number of values of N is 37 × 7 × 1 = 259.

Therefore, the answer is C.

Answer

D

Answer

The no.s of zero is determined by the no.s of 5 & 2 . which ever is less . since 5×2=10. in case the numbers of 5 is less than 2
then , N/5+N/5^2+N/5^3+N/5^4=37
the value of N = 150,151,152,153,154=5 values.

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