100 Factorial in Base 6: How Many Zeros?Date: 10/24/2001 at 15:27:22 From: Krista Subject: Factorial in base 6 How many zeros are at the end of 100! in base 6? I have tried to figure it out by writing out the entire list of base 6 numbers up to 100. I come up with 7 just using the numbers with zeroes in the end, but I have a feeling there are more. Could you please help? Thanks! Date: 10/29/2001 at 11:14:06 From: Doctor Greenie Subject: Re: Factorial in base 6 Hi, Krista - In base 10, you get a zero at the end of a product if one of the factors contains a prime factor of 5 and another of the factors contains a prime factor of 2. The prime factors of 2 and 5 multiplied together make 10 and produce a 0 at the end of the product. If you want to (for example) find the number of 0's at the end of 1000! in base 10, you need to find the total number of factors of 5 and the total number of factors of 2 in all the numbers from 1 to 1000; the smaller of those two numbers will be the number of 0's at the end of 1000! Here is how you would determine the number of 0's at the end of 1000! in base 10: 1) 1 out of every 5 numbers 1 to 1000 inclusive (200 of them) contains at least one factor of 5. 2) Of the 200 numbers 1 to 1000 inclusive that contain at least one factor of 5, 1 out of every 5 (40 of them) contains a second factor of 5. 3) Of the 40 numbers 1 to 1000 inclusive that contain a second factor of 5, 1 out of every 5 (8 of them) contains a third factor of 5. 4) Of the 8 numbers 1 to 1000 inclusive that contain a third factor of 5, 1 out of every 5 (1 of them) contains a fourth factor of 5. The total number of factors of 5 contained in the numbers 1 to 1000 inclusive is then 200 + 40 + 8 + 1 = 249 Performing the arithmetic without writing down all the words to describe what we are doing, the actual determination of the number of factors of 5 contained in the numbers 1 to 1000 inclusive looks like this: 1000/5 = 200 200/5 = 40 40/5 = 8 8/5 = 1 1/5 = 0 ---- 249 We know now that there are 249 factors of 5 contained among the numbers 1 to 1000 inclusive. There will be 249 0's at the end of 1000! if there are at least 249 factors of 2 contained among the numbers 1 to 1000 inclusive. We know that there are, because each even number (500 of them from 1 to 1000 inclusive) contains at least one factor of 2. However, to practice the method we use to solve this kind of problem, let's go ahead and find the total number of factors of 2 contained in the numbers 1 to 1000 inclusive. We use the shorthand form of the calculation as demonstrated immediately preceding this paragraph: 1000/2 = 500 500/2 = 250 250/2 = 125 125/2 = 62 62/2 = 31 31/2 = 15 15/2 = 7 7/2 = 3 3/2 = 1 1/2 = 0 ----- 994 And now on to your problem. First, I am assuming that the 100 in 100! is a base-10 number, and not a base-6 number. (If it was supposed to be a base-6 number equivalent to 36 base 10, then simply replace my starting point 100 in the calculations below with 36.) To get a 0 at the end of a product in base 6, you need prime factors that, when multiplied together, give you 6 - so you need a factor of 2 and a factor of 3. To find the number of 0's at the end of 100! in base 6, then, you need to determine the total number of factors of 2 and the total number of factors of 3 among the numbers 1 to 100 inclusive. number factors of 2 in 100!.... 100/2 = 50 50/2 = 25 25/2 = 12 12/2 = 6 6/2 = 3 3/2 = 1 1/2 = 0 ----- 97 number factors of 3 in 100!.... 100/3 = 33 33/3 = 11 11/3 = 3 3/3 = 1 1/3 = 0 ---- 48 With 48 factors of 3 and 97 factors of 2 among the numbers 1 to 100 inclusive, there will be 48 0's at the end of the base-6 representation of 100! I hope this helps. Write back if you have any further questions on this topic. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/ |
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