# CAT Practice : Number System: Factorial

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We can find the number of trailing zeroes a factorial have. Can you find a number that has a specific number of trailing zeroes? Can you give a certain number of trailing zeroes that are not present in any factorial?

## Factorials - Trailing zeroes

Q.7: Find the least number n such that no factorial has n trailing zeroes, or n + 1 trailing zeroes or n + 2 trailing zeroes.
1. 153
2. 126
3. 624
4. 18

Choice A. 153

## Detailed Solution

The previous question includes a detailed discussion on how to find the number of trailing zeroes of n!, for any natural number n.

We see that 24! has ${{{{\rm{24}}} \over {\rm{5}}}}$ = 4 zeroes
25! ends with $\left[ {{{{\rm{25}}} \over {\rm{5}}}} \right]{\rm{ + }}\left[ {{{{\rm{25}}} \over {\rm{5}}}} \right]$ = 6 zeroes. There is no natural number m such that m! has exactly 5 zeroes.

Similarly, we see that 49! ends with$\left[ {{{{\rm{49}}} \over {\rm{5}}}} \right]{\rm{ + }}\left[ {{{{\rm{49}}} \over {{\rm{25}}}}} \right]$ = 10 zeroes, whereas 50! ends with $\left[ {{{{\rm{50}}} \over {\rm{5}}}} \right]{\rm{ + }}\left[ {{{{\rm{50}}} \over {{\rm{25}}}}} \right]$ = 12 zeroes. No factorial ends with 11 zeroes.

So, any time we have a multiple of 25, we 'skip' a zero. This is because a multiple of 25 adds two zeroes to the factorial.

Extrapolating this, we can see that 125 might actually 'skip' two zeroes. 124! ends with $\left[ {{{{\rm{124}}} \over {\rm{5}}}} \right]{\rm{ + }}\left[ {{{{\rm{124}}} \over {{\rm{25}}}}} \right]$ = 24 + 4 = 28 zeros, whereas 125! has $\left[ {{{{\rm{125}}} \over {\rm{5}}}} \right]{\rm{ + }}\left[ {{{{\rm{125}}} \over {{\rm{25}}}}} \right]{\rm{ + }}\left[ {{{{\rm{125}}} \over {{\rm{125}}}}} \right]$ = 25 + 5 + 1 = 31 zeros. There is no factorial with 29 or 30 zeros.

In order to jump three zeros, think about what we need to look at. Every multiple of 25 gives us one 'skipped' zero. Every multiple of 125 gives us two 'skipped' zeroes.

In order to have three skipped zeroes, we need to look at 624! and 625!
624! has $\left[ {{{624} \over {\rm{5}}}} \right]{\rm{ + }}\left[ {{{624} \over {{\rm{25}}}}} \right]{\rm{ + }}\left[ {{{{\rm{624}}} \over {{\rm{125}}}}} \right]$ = 124 + 24 + 4 = 152 zeros
625! has $\left[ {{{625} \over {\rm{5}}}} \right]{\rm{ + }}\left[ {{{625} \over {{\rm{25}}}}} \right]{\rm{ + }}\left[ {{{{\rm{625}}} \over {{\rm{125}}}}} \right]{\rm{ + }}\left[ {{{{\rm{625}}} \over {{\rm{625}}}}} \right]$ = 125 + 25 + 5 + 1 = 156 zeros

There is no factorial with 153, 154 or 155 zeros. Or the least value of n such that no factorial ends with n, (n + 1) or (n + 2) zeroes is 153.

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