Guest Posted December 31, 2009 Report Share Posted December 31, 2009 Determine the probability that for a positive integer N, chosen at random between 1 and 9 inclusively, the number 1111*N is expressible as the difference of two positive perfect squares in at least one way. Quote Link to comment Share on other sites More sharing options...
0 Izzy Posted December 31, 2009 Report Share Posted December 31, 2009 So, randomly thinking.. at 4:30 am, forgive me. The first few perfect squares = 1,4,9,16,25. The differences of these are 3,5,7,9, which can be expressed as 2n+1. So 1111 = 2n+1, n=555. So, the 555th difference is 1111? I guess plug them all in, and it works with whole numbers, doesn't with numbers that aren't integers. 1111 = 555 2222 = 1110.5 3333 = 1666 4444 = 2221.5 5555 = 2777 6666 = 3332.5 7777 = 3888 8888 = 4443.5 9999 = 4999 (It's cool how the numbers themselves are patterns like that.) So, er. 5/9? Quote Link to comment Share on other sites More sharing options...
0 Izzy Posted December 31, 2009 Report Share Posted December 31, 2009 (edited) Actually, I think I'm right. I just used an online perfect square producer (to save me the trouble), and it works for 1111. I'd check higher.. but it won't let me go over 1000, I cba to get a calculator, and I *really* need to go to sleep. *edit* Well, I'm right for my nondecimal answers. I haven't thought of a way to check for the others yet. *edit* Google calculator ftw. I'm right about the ones I'm right about. Fairly sure about the rest. Edited December 31, 2009 by Izzy Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 31, 2009 Report Share Posted December 31, 2009 the number 1111*N expressed as difference of two positive perfect squares, which means 1111*N = a^2-b^2 = (a+b)(a-b) so a+b = 1111 a-b = N only odd numbers satisfy this condition so probability is 5/9 = 0.555555555555555.. Quote Link to comment Share on other sites More sharing options...
0 superprismatic Posted December 31, 2009 Report Share Posted December 31, 2009 The probability is 7/9. Since the difference of two squares, say a2-b2, can be factored as (a-b)*(a+b) we have that a-b must account for some of the factors of 1111*N (call their product F) and a+b must account for the rest (call their product G). So, we have to solve a-b=F a+b=G From this, we know that 2*a = F+G. So F and G must have the same parity in order for the above linear system to be solvable in the positive integers. Thus, if N has any factors of 2, it must have an even number of them -- just one (or an odd number) would force F and G to be of different parities. We exhibit one way to express 1111*N for all N for which it is possible: 1111 = 562-452 2222 can't be done: only one 2 in its factorization 3333 = 672-342 4444 = 1122-902 5555 = 782-232 6666 can't be done: only one 2 in its factorization 7777 = 892-122 8888 = 1232-792 9999 = 1682-1352 Quote Link to comment Share on other sites More sharing options...
0 Izzy Posted December 31, 2009 Report Share Posted December 31, 2009 ^Nice. Quote Link to comment Share on other sites More sharing options...
0 superprismatic Posted December 31, 2009 Report Share Posted December 31, 2009 The probability is 7/9. Since the difference of two squares, say a2-b2, can be factored as (a-b)*(a+b) we have that a-b must account for some of the factors of 1111*N (call their product F) and a+b must account for the rest (call their product G). So, we have to solve a-b=F a+b=G From this, we know that 2*a = F+G. So F and G must have the same parity in order for the above linear system to be solvable in the positive integers. Thus, if N has any factors of 2, it must have an even number of them -- just one (or an odd number) would force F and G to be of different parities. We exhibit one way to express 1111*N for all N for which it is possible: 1111 = 562-452 2222 can't be done: only one 2 in its factorization 3333 = 672-342 4444 = 1122-902 5555 = 782-232 6666 can't be done: only one 2 in its factorization 7777 = 892-122 8888 = 1232-792 9999 = 1682-1352 "Thus, if N has any factors of 2, it must have an even number of them -- just one (or an odd number) would force F and G to be of different parities." is not true. It should have read "Thus, N may not have just one factor of 2 as this would force F and G to be of different parities." Quote Link to comment Share on other sites More sharing options...
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Determine the probability that for a positive integer N, chosen at random between 1 and 9 inclusively, the number 1111*N is expressible as the difference of two positive perfect squares in at least one way.
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