Thursday, October 23, 2014

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*4^5*5^6 is 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 \(3^{4}\times 2^{10}\times 5^{6}2\) 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.
Add them up and the answer is 26.