maths: westpac maths competiton 2008 HELP!?
does anyone know how to solve this question?
let us call a number a 'cool' number if:
the sum of it starts and ends with 1 and the terms are different by at most 1
eg 1+2+3+3+4+3+2+3+3+3+2+1 is a cool number
how many terms would you need to use to add up to 2008 using as few terms as you can???
the answer is a whole number between 0 and 999 ...
THANKS!
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sum from 1 to n, with n being 1 of the 2 middle numbers
= n(n+1)/2
n(n+1)/2 ≤ 2008/4 < (n+1)(n+2)/2
n(n+1) ≤ 1004 < (n+1)(n+2)
n = 44, but since 44*45 < 1004 and not equal,
answer = 2*44 + 1 = 89
Looks like 89.
Sum 1+2+3+4+...+43+44+45+44+43+...+4+3+2+1 = 2025
is larger than 2008, so we can sum to it using 89 terms...
and 1+2+3+4+...+43+44+44+43+...+4+3+2+1 = 1980
is less than 2008, so apparently 88 terms isn't enough.
i did this too, i reckon i got 26/30 as score. anyway i think all you have to do is find a set of consecutive numbers going up and down that add to 2008.
Therefore the first set of consecutive numbers must add to 1004-x for the middle number x.
I got some answer but don't remember it.
That was a really hard paper.
lol
"cool number"
who writes this ****?