How many values of n exist such that n ends with exactly 30 zeros?

How many values of n exist such that n ends with exactly 30 zeros?

So there is no such number which ends with exactly 30 zeroes. You will see that 125! has 125/5 +125/25 +125/125 =25+5+1 =31 zeroes.

For what value of N does n terminate in 100 zeros?

For n=405 there will be 100 zeroes at the end.

What is N factorial?

In mathematics, the factorial of a non-negative integer n, denoted by n!, is the product of all positive integers less than or equal to n: For example, The value of 0! is 1, according to the convention for an empty product.

How many zeros are there at the end of 25 factorial?

25 is the square of 5 and hence it has two 5s in it. In toto, it is equivalent of having six 5s. There are at least 6 even numbers in 25! Hence, the number 25! will have 6 trailing zeroes in it.

How many zeros are there at the end of 25?

How many values of n are there such that n ends in 1998 zeros?

Therefore, all values of n are 8000,8001,8002,8003, and 8004.

What is the maximum power of 2 in 48?

16
Output : 1<<4 =16 // pow(2,4) = 16 Highest power of 2 that divides 48 is 16.

For what value of N does n terminates in 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.

What is the smallest N for which N has more than 10 digits?

The answer key says n=14.

How do you calculate N?

Factorial of a whole number ‘n’ is defined as the product of that number with every whole number till 1. For example, the factorial of 4 is 4×3×2×1, which is equal to 24….Factorial.

1.	What Is Factorial?
2.	Formula for n Factorial
3.	Factorial of Negative Numbers
4.	Use of Factorial
5.	Calculation of Factorial

What is the sum of Z trailing zeros in the infinite geometric series?

The infinite geometric series has sum n 4. So if we want z trailing zeros, then n > 4 z. A computation using 4 z will tell us how much we need to go forward from 4 z. The difference between n 5 k and ⌊ n 5 k ⌋ is always less than 1.

What is the value of n=0 + 5 a 1 + 25?

The constraint 0 ≤ a 1, a 2 ≤ 4 further implies that a 2 cannot be 0 and 1. Hence, a 2 = 2. This gives us a 1 = 1. Hence, n = a 0 + 5 a 1 + 25 a 2 = a 0 + 5 × 1 + 25 × 2 = 55 + a 0 where a 0 ∈ { 0, 1, 2, 3, 4 }.

What is the first multiple of 5 after Z = 13?

Note that the first n that works (if there is one) is a multiple of 5. And then n + 1, n + 2, n + 3, and n + 4 are the others. By that criterion, when z = 13, our first candidate is 55, the first multiple of 5 after ( 4) ( 13), and it works.