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The best strategy which i have come across is :
The highest power of prime number p in n! = [n/p1] + [n/p2] + [n/p3] + [n/p4] + …
[x] denotes the greatest integer less than or equal to x.
[1.2] =1
[4] = 4
For example: Find the highest power of 3 in 100!
= [100/3] + [100/32] + [100/33] + [100/34] + [100/35] + …
= 33 + 11 + 3 + 1 + 0 (from here on greatest integer function evaluates to zero)
= 48.
Now coming back to your first problem:
Number of 5's in 100! =?
[100/5] + [100/(5^2)] = 20+4=24
For no: which is non prime like 24,40 etc. use the below mentioned methodology with an example:
We learnt before that any non prime number can be expressed as a product of its prime factors.
We will use this concept to solve this type of questions.
What is the greatest power of 30 in 50!
30 = 2 * 3 * 5.
Find what the highest power of each prime is in the given factorial
We have 25 + 12 + 6 + 3 + 1 = 47 2s in 50!
16 + 5 + 1 = 22 3s in 50!
10 + 2 = 12 5s in 50!
We need a combination of one 2, one 3 and one 5 to get 30. In the given factorial we have only 12 5s. Hence only twelve 30s are possible. So the greatest power of 30 in 50! is 12.
Now let's solve the 3 rd problem which is Number of 14 in 100! =?
14=2*7
We have 50+ 25 + 12 + 6 + 3 + 1 = 97 2s
We have 14+2 = 16 7s
We need a combination of one 2, one 7 to get 14. In the given factorial we have only 16 7s. Hence only sixteen 14s are possible. So the greatest power of 7 in 100! is 16.
The highest power of prime number p in n! = [n/p1] + [n/p2] + [n/p3] + [n/p4] + …
[x] denotes the greatest integer less than or equal to x.
[1.2] =1
[4] = 4
For example: Find the highest power of 3 in 100!
= [100/3] + [100/32] + [100/33] + [100/34] + [100/35] + …
= 33 + 11 + 3 + 1 + 0 (from here on greatest integer function evaluates to zero)
= 48.
Now coming back to your first problem:
Number of 5's in 100! =?
[100/5] + [100/(5^2)] = 20+4=24
For no: which is non prime like 24,40 etc. use the below mentioned methodology with an example:
We learnt before that any non prime number can be expressed as a product of its prime factors.
We will use this concept to solve this type of questions.
What is the greatest power of 30 in 50!
30 = 2 * 3 * 5.
Find what the highest power of each prime is in the given factorial
We have 25 + 12 + 6 + 3 + 1 = 47 2s in 50!
16 + 5 + 1 = 22 3s in 50!
10 + 2 = 12 5s in 50!
We need a combination of one 2, one 3 and one 5 to get 30. In the given factorial we have only 12 5s. Hence only twelve 30s are possible. So the greatest power of 30 in 50! is 12.
Now let's solve the 3 rd problem which is Number of 14 in 100! =?
14=2*7
We have 50+ 25 + 12 + 6 + 3 + 1 = 97 2s
We have 14+2 = 16 7s
We need a combination of one 2, one 7 to get 14. In the given factorial we have only 16 7s. Hence only sixteen 14s are possible. So the greatest power of 7 in 100! is 16.
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