# CAT Practice : Number System: Factorial

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How many followers trail a great leader? How many zeroes trail a factorial?

## Factorials - Trailing zeroes

Q.5: How many trailing zeroes (zeroes at the end of the number) does 60! have?
1. 14
2. 12
3. 10
4. 8

Choice A. 14

## Detailed Solution

To start with, the number of trailing zeroes in the decimal representation of a number = highest power of 10 that can divide the number.

For instance,
3600 = 36 * 102
45000 = 45 * 103

In order to approach this question, let us first see the smallest factorial that ends in a zero.

1! = 1
2! = 2
3! = 6
4! = 24
5! = 120

Now, 5! ends in a zero as we have get a product of 10 when we compute 1 * 2 * 3 * 4 * 5.

10 is 2 * 5, so we get a factor of 10 every time we get a 2 and a 5 in the factorial.
So, 5! has 1 zero. The factorial that ends with 2 zeroes is 10!
15! has 3 zeroes.
20! has 4 zeroes and so on.

An extra zero is created every time a 2 and 5 combine. Every even number gives a two, while every fifth number gives us a 5.

Now, the critical point here is that since every even number contributes at least a 2 to the factorial, 2 occurs way more frequently than 5. So, in order to find the highest power of 10 that can divide a number, we need to count the highest power of 5 that can divide that number. We do not need to count the number of 2’s in the system as there will be more than 2’s than 5’s in any factorial.

Now, every multiple of 5 will add a zero to the factorial. 1 * 2 * 3 *.......59 * 60 has twelve multiples of 5. So, it looks like 60! will end in 12 zeroes. But we need to make one more adjustment here.

25 is 52, so 25 alone will contribute two 5’s, and therefore add two zeroes to the system. Likewise, any multiple of 25 will contribute an additional zero.

So, 20! has 4 zeroes, 25! has 6 zeroes.

60! will have $\left[ {{{{\rm{60}}} \over {\rm{5}}}} \right]$ zeroes arising due to the multiples of and an additional $\left[ {{{{\rm{60}}} \over {\rm{25}}}} \right]$ due to the presence of 25 and 50. {We retain only the integer component of $\left[ {{{{\rm{60}}} \over {\rm{25}}}} \right]$ as the decimal part has no value}

So, 60! will end with 12 + 2 zeros. = 14 zeros.

In general, any n! will end with $\left[ {{{\rm{n}} \over {\rm{5}}}} \right]{\rm{ + }}\left[ {{{\rm{n}} \over {{\rm{25}}}}} \right]{\rm{ + }}\left[ {{{\rm{n}} \over {{\rm{125}}}}} \right]{\rm{ + }}\left[ {{{\rm{n}} \over {{\rm{625}}}}} \right]$.........zeroes.

Generalizing further, in case we want to find the highest power of 3 that divides n!, this is nothing but $\left[ {{{\rm{n}} \over {\rm{3}}}} \right]{\rm{ + }}\left[ {{{\rm{n}} \over {\rm{9}}}} \right]{\rm{ + }}\left[ {{{\rm{n}} \over {{\rm{27}}}}} \right]{\rm{ + }}\left[ {{{\rm{n}} \over {{\rm{81}}}}} \right]$............

The highest power of 7 that divides n! is $\left[ {{{\rm{n}} \over {\rm{7}}}} \right]{\rm{ + }}\left[ {{{\rm{n}} \over {{\rm{49}}}}} \right]{\rm{ + }}\left[ {{{\rm{n}} \over {{\rm{343}}}}} \right]{\rm{ }}$...........

In case of a composite number, we need to break into the constituent primes and compute the highest power that divides the number.

For instance, if we want to find the largest power of 15 that divides n!, this will be driven by the highest powers of 3 and 5 that divide n!. Similar to the scenario we saw with trailing zeroes, we can observe that there will definitely be at least as many 3’s than 5’s in any factorial. So, the highest power of 15 that divides n! is simply $\left[ {{{\rm{n}} \over {\rm{5}}}} \right]{\rm{ + }}\left[ {{{\rm{n}} \over {{\rm{25}}}}} \right]{\rm{ + }}\left[ {{{\rm{n}} \over {{\rm{125}}}}} \right]{\rm{ + }}\left[ {{{\rm{n}} \over {{\rm{625}}}}} \right]$............

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