Posted by Lucy on Tuesday, April 9, 2013 at 10:37am.
How many numbers
from 1 to 1000
inclusive can be
expressed as the sum
of k >= 2 consecutive
positive integers for
some value of k ? Sorry to post question from Brilliant, I got 997 integers, but it's wrong. If you think you can answer ...
Math ( Number Theory ) - Steve, Tuesday, April 9, 2013 at 11:11am
For k=2, we want all n<=1000 such that n=m+m+1 = 2m+1
That is, all odd numbers 3<=n<=999
There are 996/2+1 = 499 of those
For k=3, we want n=3m+3
so, that's all multiples of 3 with 6<=n<=999
There are 993/3+1 = 332 of those
For k=4, we want 4m+6
so, that's all multiples of 4 with 10<=n<=998
There are 988/4+1 = 248 of those
There's a pattern here, so you should be able to come up with some
expression, or just brute-force your way through.
Maybe not a lot of help, but it does show that there are more than 997 of them.
Math ( Number Theory ) - Arpan, Wednesday, April 10, 2013 at 9:12am
The ans Steve gave is absolutely wrong...the ans is less than 823 and greater than 665...steve u have counted the same numbers more than one time...for example u have counted 15 twice....
Math ( Number Theory ) - Anonymous, Wednesday, April 10, 2013 at 4:49pm
When I calculated out I got 990 as the correct answer.
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