Options are:

  1. 20
  2. 21
  3. 22
  4. 23
  5. 24

I recently came across the above question in a competitive exam, where we get about 30 seconds to 1 minute for solving each problem. I want to if there are quick and easy methods to compute the result.

  • 2
    $\begingroup$ HINT: The number of $0$'s in a number is equal to the number of times $10$ divides it. $\endgroup$ – Marcus M Sep 15 '15 at 18:22
  • $\begingroup$ @MarcusM That is not true: consider, for instance, $101$. $\endgroup$ – Théophile Sep 15 '15 at 18:33
  • $\begingroup$ I think Marcus may have chosen a more accurate duplicate, in hindsight, but I'll leave mine up, just in case. $\endgroup$ – Cameron Buie Sep 15 '15 at 18:40

As a general rule, the number of $0$s at the end of $k!$ is equal to $$\sum_{i=1}^\infty \left\lfloor\frac{k}{5^i}\right\rfloor$$

The above expression is a measure of how many times $5$ divides $k!$.


A zero can only occur as result of multiplying 5 with 2. so consider the factors 5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90,95,100 now Tolal number of 5 that can be obtained from these factor are 24.so there wiil be 24 zero.

  • $\begingroup$ Instead of writing a brand new answer, it is better practice to fix the one that's being downvoted. $\endgroup$ – Cameron Buie Sep 15 '15 at 18:36
  • $\begingroup$ @CameronBuie,thanks for your suggestion. $\endgroup$ – Manoj upreti Sep 15 '15 at 18:39
  • $\begingroup$ @michael Zero can also occur by multiplying 0 by 0(10x10=100). I didn't quiet get the solution, could you please elaborate. $\endgroup$ – Aura Wanderer Sep 15 '15 at 18:39
  • $\begingroup$ @AuraWanderer,since if we have a number ending with say one zero than it must be result of presence of 5 and 2 in its factor like 20=5*2*2,so here we have a pair (5,2).similarly for number ending with two zero say 200=5*5*2*2*2 so we have two five's here which after multiplication with 2 will make out two zero.hope this will help. $\endgroup$ – Manoj upreti Sep 15 '15 at 18:46

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