Sums of Consecutive IntegersDate: 08/18/99 at 23:12:54 From: Eric Subject: Adding consecutive numbers Is there a way of finding a series of consecutive integers whose sum is a given number? For example: 3 = 1 + 2 (There is no series that produces a sum of 4) 5 = 2 + 3 6 = 1 + 2 + 3 7 = 3 + 4 (Is there one that produces a sum of 8?) 9 = 2 + 3 + 4, and 9 = 4 + 5 and so on... Thanks. Date: 08/19/99 at 16:56:07 From: Doctor Rick Subject: Re: Adding consecutive numbers Hi, Eric. You've asked an interesting question. Are you investigating this on your own, or is it a school project? You have discovered the beginning of a pattern. To learn more than this, though, takes algebra. It took a little work, but I believe I have proved that this can be done for any number that has an odd factor (other than 1). In other words, if there is an odd number other than 1 that divides evenly into the number, then there is a list of consecutive numbers that add to the number. The number 4 has no odd factors, since 4 = 2 x 2. The same is true for 8, since 8 = 2 x 2 x 2. Numbers like this (called powers of 2) are the only numbers that will not work; all others will. The next number that won't work is 16. Here is a method for finding the series of consecutive numbers that add to a particular number. I'll demonstrate it with the number 52. Find an odd factor: 52 is divisible by 13. Double the original number and divide it by the factor you chose: 2 * 52 / 13 = 8 Subtract 8 from 13 (the smaller number from the bigger number) to get 5; add 1 and divide by 2 to get 3. Start with 3 and write 8 numbers (the smaller of the 2 factors), and this series will add to 52. 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 52 Sometimes, as you found, there is more than one solution. Any number that has more than one odd prime factor will have more than one solution. For a sum of 9, we look for odd factors, and we find 3 and 9. I'll start with 3. 3 goes into 18 6 times; subtract 3 from 6, add 1, and divide by 2 to get 2. Write 3 numbers starting with 2: 2 + 3 + 4 = 9 Using the other odd factor, 9, we divide 18 by 9 to get 2. Subtract 2 from 9, add 1 and divide by 2 to get 4. Write 2 numbers starting with 4: 4 + 5 = 9 just as you found. The way I've described it, this is just magic. But believe me, I found the formula using math. If you want to learn more about how I did it, write back. I'd be glad to explore it with you. In the meantime, you have a nice little trick to show people. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ Date: 01/27/2002 at 19:24:34 From: R. Lineage Subject: Summing Consecutive Numbers Hi, Has anyone written back? I played around with it a bit, and noticed that if the number is twice a prime (e.g. 46), then you have to use Dr. Rick's method. But if it's not twice a prime, i.e. 294, then there's a slightly easier way. Find any odd prime factor (the smaller the better) and divide this into the number. [3 is an odd prime factor of 294, and 294/3 = 98.] The factor (3) is the number of terms needed. The answer (98) is the middle number. So, in this case, 294 = 97+98+99 You could do it with the factor of 7 as well. In this case, 294/7 = 42, so there would be 7 terms, with the middle term being 42. 294 = 39+40+41+42+43+44+45 This is a bit easier than Dr Rick's method. But I still would like to know how he came up with his method. Thanks. Date: 01/28/2002 at 14:24:51 From: Doctor Rick Subject: Re: Summing Consecutive Numbers Hi, R. Unfortunately, Eric did not write back, and it's been almost 2 1/2 years since I worked out the formula, so I don't recall exactly what I did! I'll try to reconstruct my work. If we have a sequence of k consecutive integers starting with n, then their sum is n + (n+1) + (n+2) + ... + (n+k-1) = kn + k(k-1)/2 = k(n+(k-1)/2) = k(2n+k-1)/2 Here I have used the formula for the sum of (k-1) consecutive integers beginning with 1. The sum must be the number we are seeking to write as a sum of consecutive integers; I'll call it S. S = k(k+2n-1)/2 In order for S to be an integer, one of the two factors k and (k+2n-1) must be even. In fact, since their difference (2n-1) is odd, exactly one of these numbers is even, and the other is odd. In either case, we can solve for n in terms of k: k+2n-1 = 2S/k n = (2S/k - k + 1)/2 If k is an odd factor of S, then (k+1)/2 is an integer, and (2S/k)/2 = S/k is an integer, therefore n is an integer. If k is even, then (k+2n-1) is an odd factor of S. Call this factor f, and we find S = kf/2 k = 2S/f n = (2S/k - k + 1)/2 = (f - 2S/f + 1)/2 Thus we have two solutions for each odd factor of S: n = (2S/f - f + 1)/2, k = f n = (f - 2S/f + 1)/2, k = 2S/f These solutions are related: for example, for S = 68 = 2*2*17, the solutions are (with f = 17, the only odd factor) n = (2*68/17-17+1)/2 = -4, k = 17 -4+-3+-2+-1+0+1+2+3+4+5+6+7+8+9+10+11+12 = 68 n = (17-2*68/17+1)/2 = 5, k = 2*68/17 = 8 5+6+7+8+9+10+11+12 = 68 If n < 0 and k > 2n+1, then the first 2n+1 terms total zero and can be eliminated, leaving the other solution. This was the solution I gave in words: Double the original number [S] and divide it by the odd factor you chose [K, obtaining 2S/K]. Subtract the smaller factor [k] from the greater factor; add 1 and divide by 2. Start with this number [n] and write [k] numbers, and this series will add to [S]. This solution ensures that we get the solution with positive n. Now let's consider your simplification. If the number S is not twice a prime, then your formula is to pick a (small) odd prime factor of S, use this as k (the number of consecutive numbers) and use 2S/k as the middle term. The middle term (if k is odd, so there is a middle term) is n+(k+1)/2. My first formula for n makes the middle term n+(k-1)/2 = (2S/k-k+1)/2 + (k+1)/2 = S/k The only difference between our formulas, therefore, is that you only get one of the two solutions; in some cases it is the solution with negative starting number. For instance, for S = 2*17 = 34, you get 17 terms with the middle number 2, or -6+-5+-4+-3+-2+-1+0+1+2+3+4+5+6+7+8+9+10 = 34 which is a valid solution, but it can be reduced to 7+8+9+10 = 34 Let's combine our two solutions: To find a sequence of consecutive integers that add to a given positive integer S, choose any odd factor of S, and call this factor f. (If S is a power of 2 and thus has no odd factor, there is no solution.) Write down f consecutive integers with S/f as the center number (thus, the first number is S/f-(f-1)/2, and the last number is S/f+(f-1)/2). This sequence adds to S. If the first number is less than zero, drop the first numbers through the negative of the first number; the sum will be the same. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ Date: 01/28/2002 at 17:11:51 From: R. Lineage Subject: Summing Consecutive Numbers Thanks, Dr Rick, for your very quick reply. I find it intriguing how mathematicians come up with stuff. I am just beginning maths teaching in Australia. I love maths, but I don't think I'm what they'd call a natural. But I get a major kick with number theory. E.g. when I found out about wondrous numbers, and the challenge to see if 27 was a wondrous number, I couldn't help myself. I set up an Excel spreadsheet to do a calculation for all numbers from 1 to 1000, and all of them eventually went to 1. The formulae were actually pretty simple, but it was only then that I was happy! Anyway, I think the whole Dr Math site is absolutely fantastic. I'm not interested in teaching maths if I can't explain things clearly to my students, and the site is helping me lots. Thanks again... |
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