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Highest Powers of a Number Contained in a Factorial

Rule1 : To find powers of a prime number p contained in n! where n is a natural number we proceed in this manner:

Highest power of prime number p in n! = [n/p] + [n/p ² ] + [n/p ³ ] + [n/p ⁴ ] + ...

where

• denotes the greatest integer less than or equal to x.

Rule2 : to find the highest power of a composite number C in n! where n is a natural number we proceed in this manner.

Let C = (x) ^a (y) ^b (z) ^c .. where x, y, z, etc. are prime factors of C. We find the largest prime factor of C and find its highest power contained in n! by rule 1. The highest power of C in n! Is the highest power of the largest prime factor of C in n!.

EXAMPLES

26. Find the highest power of 2 in 50!.

The highest power of 2 in 50! = [50/2] + [50/4] + [50/8] + [50/16] + [50/32]

= 25 + 12 + 6 + 3 + 1 = 47
Find the highest power of 30 in 50!

30 = 2 x 3 x 5. Now 5 is the largest prime factor of 30 so we find the highest power of 5 in 50!.

The highest power of 5 in 50! = [50/5] + [50/25] = 10 + 2 = 12

Hence the highest power of 30 in 50! = 12
Find the number of zeroes present at the end of 100!.

We get a zero at the end of a number when we multiply that number by 10. So, to calculate the number of zeroes at the end of 100!, we have to find the highest power of 10 present in the number. Since 10 = 2 x 5, we have to find the highest power of 5 in 100!

The highest power of 5 in 100! = [100/5] + [100/25] = 20 + 4 = 24

So, the number of zeroes at the end of 100! = 24

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