Squares in an Infinite Factorial SeriesDate: 11/23/2001 at 10:32:31 From: Gaurav Tekriwal Subject: Squares in an Infinite Factorial Series How many perfect squares appear among the following numbers? 1!, 1!+2!, 1!+2!+3!,...1!+2!+3!+...n! My approach was looking for patterns or cycles as such, but I wasn't able to find any. Is there an interesting way to find the number of perfect squares? I also think that the number would tend to infinity. Also, if we bound the sequence to a limit, say k!, then is it possible to find the number of perfect squares? Thanks, GAuRAv Date: 11/23/2001 at 11:07:25 From: Doctor Tom Subject: Re: Squares in an Infinite Factorial Series Hi Gaurav, You didn't look too carefully: 1! + 2! + 3! = 1 + 2 + 6 = 9. I'm not sure how to prove whether or not there are an infinite number, but there's at least one. - Doctor Tom, The Math Forum http://mathforum.org/dr.math/ Date: 11/23/2001 at 12:42:31 From: Doctor Jubal Subject: Re: Squares in an Infinite Factorial Series Hi Gaurav, The solution that Dr. Tom found, 9 = 3^2 = 1!+2!+3!, along with 1 = 1^2 = 1!, are the only solutions. Here's how you can prove this is true. Let's look at the next largest term in the sequence 1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33 Note the last digit of 33 is 3. Now, any factorial greater than 4! = 24 is divisible by 10, because they have the form 1*2*3*4*5*(more stuff), and so are divisible by both 2 and 5. Thus, the last digit of all factorials beginning with 5! = 120, is 0. No matter how many times we add zero to 3, we still get 3, so the last digit in every one of the sums of the factorials you're interested in, beginning with 33, is 3. However, no perfect square has 3 as its last digit. To prove this, observe that we can write any integer as n = 10a + b, where b is the last digit and a is the rest of the digits. Then n^2 = (10a + b)^2 = 100a^2 + 20ab + b^2 100a^2 and 20ab both are divisible by 10 and so have zero as the last digit, so the last digit of n^2 depends only on b^2. Thus, the last digit of any integer squared is the same as the square of the number's last digit. Looking at the ten digits 0^2 = 0 1^2 = 1 2^2 = 4 3^2 = 9 4^2 = 16 5^2 = 25 6^2 = 36 7^2 = 49 8^2 = 64 9^2 = 81 None of them has 3 as the last digit of its square, so no number has three as the last digit of its square. All the factorial sums you're looking at that are greater than 9 end in three, so none of them can be a perfect square. Does this help? Write back if you'd like to talk about this some more, or if you have any other questions. - Doctor Jubal, The Math Forum http://mathforum.org/dr.math/ |
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