# How many zeros are at the end of 300! ?

300! is a rather large number, equal to 300×299x298x…x3×2x1, but the number of trailing zeros is a simple calculation that you should be able to do in your head. Enjoy.

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Two?

rebbel (35553)

Edited because I think I got it wrong.

Mimishu1995 (23644)

From 290 to 210 we have nine 0. Now multiply 9 with 3 (from 190 to 110 and from 90 to 10) we have 27.

300, 200 and 100 give us six more 0. Now we have thirty three 0 in total.

Is that the correct answer? I’m horrible at math so I’m not so sure…

Mimishu1995 (23644)

300 factorial is an enormous number. Since it will necessarily be displayed in scientific notation there will not be any zeros at the end.

@Mimishu1995 got it.

Each factor that is a multiple of 10 or 100 adds an order or two of magnitude. 300! is 33 orders of magnitude greater than 9!.

zenvelo (39467)

You are forgetting something. To get a zero you need a factor of 10, and to get a 10 you need to pair a 5 and a 2. The 5 and the 2 can come from anywhere, for example the 5 in 55 can be paired with one of the 2’s in 4.

Like this ? 74 zeros

300! Factorial of 300

Value of 300!, i.e. factorial of 300 =
306057512216
440636035370
461297268629
388588804173
576999416776
741259476533
176716867465
515291422477
573349939147
888701726368
864263907759
003154226842
927906974559
841225476930
271954604008
012215776252
176854255965
356903506788
725264321896
264299365204
576448830388
909753943489
625436053225
980776521270
822437639449
120128678675
368305712293
681943649956
460498166450
227716500185
176546469340
112226034729
724066333258
583506870150
169794168850
353752137554
910289126407
157154830282
284937952636
580145235233
156936482233
436799254594
095276820608
062232812387
383880817049
600000000000
000000000000
000000000000
000000000000
000000000000
000000000000
000

74

300 divided by 5 =60
60 divided by 5 = 12 (round down to 10)
10 divided by 5 = 2
60+ 12+2= 74

Lightlyseared (34679)

Yes, 74 is the correct answer. Let me expand on what @Lightlyseared wrote,

We want to pair each factor of 5 with a factor of 2. It should be apparent that there are many more 2’s than 5’s, so we are limited by the number of 5’s. We can just count 5’s, knowing there will be a sufficient number of 2’s to pair with.

Every fifth number is divisible by 5, so there are 300/5=60 numbers with at least one factor of 5. To find how many additional 5’s we get from numbers divisible by 25, we can divide 300 by 25 or, more simply, divide 60 by 5, to get 12. For numbers that contribute yet another 5, we can divide 300 by 125 or 12 by 5 to get the 2 numbers 125 and 250. 60 + 12 + 2= 74.

74

I do these calculations in my head everyday while designing new rockets.

Forever_Free (8665)

I did not think this type of problem had any practical applications. How does it relate to rocket design?

Forever_Free (8665)