Puzzle: Consecutive Integers

1. mathmom permalink

   October 12, 2009 pm31 10:15 pm 10:15 pm

   If we’re counting only positive integers (which I know, we’re not) I think there are only 3 (or 4 if you count the degenerate case of expressing 1000 as the sum of one positive integer), one for each of the odd factors of 1000.

   For each of those 4, I can think of a way of also creating a sum of consecutive integers that includes negative integers.

   I think maybe that’s it, so 7 or 8 depending on whether the sum with one term counts or not.

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2. mathmom permalink

   October 12, 2009 pm31 10:20 pm 10:20 pm

   Oh, wait, one of my ways for only positive integers can’t actually be done with only positive integers, it already goes into negatives. So now I have only 6 or 7 ways.

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3. Nick permalink

   October 13, 2009 pm31 9:04 pm 9:04 pm

   Thanks for the puzzle; I count 7 solutions.

   Reply
4. Alex permalink

   October 14, 2009 pm31 4:19 pm 4:19 pm

   PART 1: odds
   ————
   Which numbers can be written as the sum of 5 consecutive integers?
   Well, we have (m-2) + (m-1) + m + (m+1) + (m+2) = 5m for all such sequences –
   It’s clear that a number can be written like this IFF it has a factor of 5.

   In general, by the same argument, a number can be written as a sum of (2k+1) integers IFF it has a factor of (2k+1).
   -> There’s one of these ways for each odd factor of N.

   PART 2: evens
   ————-
   Which numbers can be written as the sum of 100 consecutive integers? Let the first be a, and the last be b.
   a + (a+1) + (a+2) + …… + (b-2) + (b-1) + b = (a+b)*50 = (2a+99)*50.
   A number N can be written (2a+99)*50 iff N/50-99 is an even number, i.e. IFF N/50 is odd.

   In general, by the same argument, a number can be written as a sum of (2k) integers IFF N/k is an odd integer.
   -> There’s one of these ways for each odd factor of N.

   CONCLUSION
   ———-
   -> There are two unique ways for each odd factor of N
   1000 has 4 odd factors
   -> 8 ways total.

   gmail, alexrdavies1

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5. jd2718 permalink*

   October 17, 2009 am31 10:53 am 10:53 am

   This, for me, is a special puzzle.

   The first presentation I ever made to a group of math teachers, me and Mike (my equally inexperienced co-presenters) spoke on “Problem Solving” and used this as our warm-up. Actually, I don’t recall if we used 1000 or 10,000… but no real difference. The audience was full of relatively senior teachers, mostly because none of the other workshops looked so hot. And this puzzle engaged our audience and we won our audience, and the rest was easy.

   Afterwards I was approached by the chair of (at the time) the best high school in the borough, and asked if I wouldn’t like to apply for a job there…

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6. jd2718 permalink*

   October 17, 2009 am31 11:14 am 11:14 am

   The sum of consecutive integers will be the average times the number of integers… call it an

   The number of integers, n, must be an integer.

   The average,a, may be an integer, or an odd integer divided by two.

   If the average,a, is an odd integer divided by two, then the number of integers, n, must be even.

   If the average,a, is an integer, then the number of integers, n, must be odd.

   So one of the two must be either odd, or an odd number divided by two.

   Consider the factor pairs for 1000 that include an odd integer:

   1000*1, 200*5, 40*25, 8*125

   Divide the odd factors by 2:

   2000 * 1/2, 400 * 5/2, 80 * 25/2, 16 * 125/2

   You should be able to find the consecutive integers that correspond to each one of these 8 pairs of numbers.

   Reply

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