## Monday, November 7, 2016

### How Many Zeros?

Problems: (Solutions below.)

#1. 2003 Chapter Team # 7--How many zeros are at the end of (100!)(200!)(300!) when multiplied out?

#2. How many zeros are at the end of 2013!?

#3. How many zeros are at the end of 10! *9!*8!*7!*6!*5!*4!*3!*2!*1!*0!?

#4. What is the unit digit of 10! + 9! + 8! + 7! + 6! + 5! + 4! + 3! + 2! + 1! + 0!?

#5. The number $$3^{4}\times 4^{5}\times 5^{6}$$ written out in full. How many zeros are there are at the end of the number?

#6. How many zeros are at the end of (31!)/(16!*8!*4!*2!*1!)

#7. 2009 National Sprint #18-- What is the largest integer n such that $$3^{n}$$ is a factor of 1×3×5×…×97×99?

Solutions:
#1.For 100!, there are -- 100/5 = 20 , 20/5 = 4 (Stop when the quotient is not divisible by 5 and then add up all the quotients.), or 20 + 4 = 24 zeros.
For 200!, there are 200/5 =40 , 40/5 = 8, and  8/5 = 1, or total 40 + 8 + 1 = 49 zeros.
For 300!, there are 300/5 = 60, 60/5 = 12, and 12/5 = 2, or total 60 + 12 + 2 = 74 zeros.
Add all the quotients and you get 147 zeros.

#2. Use the same method as #1 and the answer is 501 zeros.

#3:Starting at 5!, you have one "0", the same goes with 6!, 7!, 8!, and 9!
10! will give you 2 extra "0"s. Thus total 7 zeros.

#4: Since starting with 5! you have "0" for the unit digit, you only need to check 4! + 3! + 2! + 1! + 0!.
24 + 6 + 2 + 1 + 1 = 34 so the unit digit is 4.

#5: Make sure to prime factorize all the given number sequences, in this case, it's $$3^{4}\times 2^{10}\times 5^{6}$$ after you do that.
2 * 5 = 10 will give you a zero since there are fewer 5s than 2s so the answer is 6 zeros.

#6: You need the same number of 2 and 5 multiple together to get a "0".
31! gives you 30/5 = 6, 6/5 = 1 or 6 + 1 = 7 multiples of 5
31! gives you 10//2 = 15...15/2 = 7...7/2 = 3...3/2 = 1 or 15 + 7 + 3 + 1 = 26 multiples of 2. 16!*8!*4!*2!*1! gives you 4 multiples of 5 and 8 + 4 + 2 + 1 (16!) + 4 + 2 + 1 (8!) + 2 + 1 (4!) + 1(2!)
= 26 multiples of 2.
Thus the multiples of 2s all cancel out, the answer is "0" zeros.

#7: There are 3*1, 3*3, 3*5...3*33 or $$\dfrac {33-1} {2}+1=17$$ multiples of 3.
There are 9*1, 9* 3...9*11 or $$\dfrac {11-1} {2}+1 = 6$$ extra multiples of 3.
There are 27*1, 27*3 or 2 extra multiples of 3.
There is 81*1 or 1 extra multiples of 3.

1. For #1 : Why do you add all the quotients together?

2. That is how you find "how many multiples of 5" because you need 5 x 2 to get a "zero" and for factorials, usually there are more 2s than 5s, so finding how many multiples of 5s is sufficient.

For example: 10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
Do you see there are 10/5 = 2, two multiples of 5 there ? which are from 5 and 10 (5x2).
However, if it's 25!, besides 25/5 = 5 multiples of 5 (5, 10, 15, 20, 25), you have an extra multiple of 5 from 5^2 or 25, which is 5 x 5.
Thus, it's 5 + 1 = 6 multiples of 5 there.
Let me know if you still have questions.
Best, Mrs. Lin