Sums of Consecutive NumbersDate: 06/20/2002 at 06:42:20 From: Laura Tyers Subject: consecutive numbers In what way can 1000 be expressed as the sum of consecutive numbers? Date: 06/20/2002 at 16:27:44 From: Doctor Ian Subject: Re: consecutive numbers Hi Laura, Well, it can't be the sum of two consecutive numbers, since that would have to be an odd number. But what about three? If the middle number is n, then (n-1) + n + (n+1) = 1000 3n = 1000 So that won't work. How about four? (n-1) + n + (n+1) + (n+2) = 1000 4n + 2 = 1000 That's not going to work, either. (Do you see why?) What didn't work for 3 will work for 5, however: (n-2) + (n-1) + n + (n+1) + (n+2) = 1000 5n = 1000 n = 200 So we've found one way. What you need to do now is figure out whether what worked for 5 will work for any other numbers! The key was to have an odd number that evenly divided 1000, so you could get a bunch of differences from the middle number to cancel out. Are there odd numbers other than 5 that evenly divide 1000? And then you're half done, because you still have to worry about the even numbers. Let's say that you want to have k consecutive numbers, where k is even. (We've tried k=2 and k=4 so far.) If we try to distribute the resulting consecutive numbers around some 'center' number, it's never quite balanced, the way it is when we use odd numbers. Let's look at some examples: k = 2 n + (n+1) k = 4 (n-1) + n + (n+1) + (n+2) k = 6 (n-2) + (n-1) + n + (n+1) + (n+2) + (n+3) Already we can see a pattern here. All the differences except the last one will cancel out, leaving us with k*n + k/2 So we can have an even number of consecutive numbers only if we can find some value of k such that k*n + k/2 = 1000 If you can find integer values of k and n that make the equation true, you'll have found more ways to add consecutive numbers to get 1000. If you can show that there are _no_ integer values of k and n that make the equation true, then you've shown that you can ignore even values of k. Can you take it from here? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ Date: 06/20/2002 at 17:21:08 From: Doctor Greenie Subject: Re: consecutive numbers Whenever you add an ODD number of consecutive integers, the total is equal to the number of numbers times the number "in the middle". For example, the sum of the three numbers 73, 74, and 75, is just 3 times 74; or the sum of the seven numbers 124, 125, 126, 127, 128, 129, and 130 is just 7 times 127. And whenever you add an EVEN number of consecutive integers, the total is equal to the number of numbers times the number halfway between the two numbers "closest to the middle". For example, the sum of the four numbers 11, 12, 13, and 14, is just 4 times 12.5; or the sum of the eight numbers 87, 88, 89, 90, 91, 92, 93, and 94 is just 8 times 90.5. So the sum of ANY sequence of consecutive integers is either (1) an ODD number times ANY INTEGER; or (2) an EVEN number times the number halfway between two integers You want to write 1000 as the sum of consecutive numbers. Let's look at the divisors of 1000: 1 1000 2 500 4 250 5 200 8 125 10 100 20 50 25 40 Let's first look at the ways we can write 1000 as the sum of an ODD number of consecutive integers. For these we need a factorization of 1000 in which one of the factors is ODD. Ignoring the divisor 1 (we aren't interested in the "sequence" consisting of the single number "1000"), there are three divisors of 1000 which are odd. We have (1) 5 x 200 = 1000 (2) 25 x 40 = 1000 (3) 125 x 8 = 1000 These factorizations tell us that we can write 1000 as (1) 5 consecutive integers with middle number 200 (2) 25 consecutive integers with middle number 40 (3) 125 consecutive integers with middle number 8 The first solution has middle number 200, with (5-1)/2=2 numbers each side of 200: 198, 199, 200, 201, 202. The second solution has middle number 40, with (25-1)/2=12 numbers each side of 40: 28, ..., 40, ... 52 The third solution has middle number 8, with (125-1)/2=62 numbers each side of 8: -54, ..., 8, ..., 70 Now let's look for the ways we can write 1000 as the sum of an EVEN number of consecutive integers. To find these solutions, we need to be able to write 1000 as the product of an even integer "a" and some number of the form "b and 1/2" where b is an integer. Any number of the form "b and 1/2" is half of an odd integer. Because the only odd divisors of 1000 are 5, 25, and 125, we have the following for the possibilities for making 1000 as the sum of an EVEN number of consecutive integers: (1) 400 x 2.5 (400 consecutive integers with middle numbers 2,3) (2) 80 x 12.5 (80 consecutive integers with middle numbers 12,13) (3) 16 x 62.5 (16 consecutive integers with middle numbers 62,63) The first solution here has middle numbers 2 and 3 with (400-2)/2=199 numbers each side of those two: -197, ..., 2, 3, ..., 202 The second solution here has middle numbers 12 and 13 with (80- 2)/2=39 numbers each side of those two: -27, ..., 12, 13, ..., 52 The third solution here has middle numbers 62 and 63 with (16-2)/2=7 numbers each side of those two: 55, ..., 62, 63, ..., 70 Notice that some of these solutions include negative integers. You didn't specify whether that was allowed. For any solution you have containing only positive integers, you will always have a corresponding solution involving negative integers, because, for example, if you have found 3+4+5=12 as one solution for a sum of consecutive integers equalling 12, then the sequence (-2)+(-1)+0+1+2+3+4+5 will also equal 12, because the (-2) and the 2 add to zero, as do the (-1) and the 1, and so the part of the sequence (-2)+(-1)+0+1+2 has the value 0. So we have three pairs of solutions to your problem (each pair consisting of a sequence of positive integers and the corresponding sequence which involves negative integers also), as follows: (1a) 198, 199, 200, 201, 202; or (1b) -197, -196, ..., 196, 197, 198, 199, 200, 201, 202 (2a) 55, 56, ..., 69, 70; or (2b) -54, -53, ..., 53, 54, 55, 56, ..., 69, 70 (3a) 28, 29, ..., 51, 52; or (3b) -27, -26, ..., 26, 27, 28, 29, ..., 51, 52 I hope you were able to follow all of this. Please write back if you have any further questions on any of this. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/ |
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