# 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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## 12 Answers

Edited because I think I got it wrong.

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…

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!.

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

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.

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

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