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1232
7
avatar+425 

Help me in Math Olympiad, the contest is on April 6!!!!

How many consecutive zeroes are at the end of:

 

a) 10!

b) 14!

c) 20!

d) 25!

 Apr 2, 2016

Best Answer 

 #4
avatar
+15

MELODY:

25! ends in 6 trailing zeros.155112 1004333098 5984,000,000

25/5=5 and 25/25=1

5  +  1  =6 zeros

 Apr 2, 2016
 #1
avatar
+3

Why don't you use a calculator?

 Apr 2, 2016
 #6
avatar+425 
0

The calculator will show up results like:

 

x × 10y or some errors that go above 9.999999999999999 × 1099

 

Also calculators are NOT allowed in the contest of Math Olympiad.

MWizard2k04  Apr 3, 2016
 #7
avatar+425 
0

When I try using a calculator, my calculator calculates only ranges of:

 

± 1 × 10-99 ≤ x < 1100 or 0.

 

The maximum factorial used for my calculator is:

 

69 !

MWizard2k04  Apr 5, 2016
 #2
avatar+118608 
0

a) 10!

 

1*2*3*4*5*6*7*8*9*10

1*2*3*4*5*6*7*8*9

one

1*3*4*6*7*8*9

two

 

1*3 ends in 3

3*4 ends in 2

2*6 ends in 2

2*7 ends in 4

4*8 ends in 2

2*9ends in 8

so the last 3 digits are 800

that is 2 zeros at the end.

 

 

 

 

b) 14!

10!  ends in 800

11! ends in 800

12! ends in 600

13! ends in 800

14! ends in 200

 

c) 20!    

15! ends in 000

20! ends in 0000        4 zeros

 

 

d) 25!

1*2*3*4*5 has 0ne zero at the end

so  add another zero

25!                           5 zeros

 Apr 2, 2016
 #3
avatar
+10

Here is a general rule to find number of trailing zeros in a factorial:

DIVIDE THE NUMBER BY POWERS OF 5.

Example: 100!

100/5^1 +100/5^2 +100/5^3.........etc

20           +  4             +    0      =24 zeros in 100 factorial. Disregard any fractions. Just take the integer part.

 Apr 2, 2016
 #4
avatar
+15
Best Answer

MELODY:

25! ends in 6 trailing zeros.155112 1004333098 5984,000,000

25/5=5 and 25/25=1

5  +  1  =6 zeros

Guest Apr 2, 2016
 #5
avatar+118608 
+5

ok I stand corrected.  Thanks :)

Melody  Apr 2, 2016

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