The integer 30 can be written as a sum of consecutive positive integers in three ways:
30 = 9+10+11 = 6+7+8+9 = 4+5+6+7+8.
Find the smallest positive integer which can be written as a sum of consecutive positive integers in 12 ways.
I am unsure about how to do this in the reverse way, but here is how this works in the forward way.
First, here is how the sums can be broken down: Since each sum is the sum of consecutive integers, we can make them all the same number by looking at the middle number. (For an even number of numbers, you clump the sets of two together before you begin.) So, this is really a factoring exercise in disguise because the "clumped" sets are all the same number added an integer number of times. The factors of 945 are 1, 3, 5, 7, 9, 15, 21, 27 | 35, 45, 63, 105, 135, 189, 315, 945.
The "positive integers" part comes into effect because the high factor (the middle number of the series) must be greater than half the low factor. (the number of terms in the series)
Also, by "unclumping" the middle number in odd series, we can create more factor chains. However, we can't count 1 and 945 since 945 by itself isn't a sequence. Also, "unclumping" when twice the low factor is greater than or equal to the high factor doesn't work. (This means you can "unclump" 21 because 44 is less than 45, but you can't can't "unclump" 27 because 54 is greater than 35. See more about this on the bottom.)
So the factor groups are: (This also counts "unclumped" versions with a = between their two middle numbers)
2-472=473
3-315, 6-157=158
5-189, 10-94=95
7-135, 14-67=68
9-105, 18-52=53
15-65, 30-31=32
21-45, 42-22=23
27-35
35-27
Because you can unclump the middle term only when there will be enough room for it to "spread out" (when doing so won't let the smallest term be less than 1), you can do so when the larger term is more than twice the smaller term. For example, 55 is equal to the sums of the numbers 9 to 13, (5-11) but is also equal to the numbers from 1 to 10 (10-5=6)
|
Posted by Gamer
on 2004-06-28 22:53:45 |