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BNC
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Sum of Consecutive numbers
« on: Jan 26th, 2003, 4:06am » |
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Every number between 1000 and 2000 may be written as the sum of Consecutive numbers, except one.
What is this number?
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| « Last Edit: Jan 26th, 2003, 6:33am by BNC » |
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How about supercalifragilisticexpialidociouspuzzler [Towr, 2007]
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Pietro K.C.
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Re: Sum of Consecutive numbers
« Reply #1 on: Jan 26th, 2003, 7:11am » |
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I like this one!
The sum of N consecutive numbers is odd, if N is even, and it is divisible by N if N is odd. The latter is simple to prove using congruences, and the first is obvious.
Now, a number that is not odd and not divisible by any odd numbers would certainly fill the prescription. What could it be? Why, a power of 2! Hence, 1024 is our number.
// solution hidden by moderator 8:26 PM 1/27/2003
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| « Last Edit: Jan 27th, 2003, 8:26pm by william wu » |
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towr
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Re: Sum of Consecutive numbers
« Reply #2 on: Jan 26th, 2003, 7:19am » |
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sum(i, i, n, m) = sum(i, i, 0, m) - sum(i,i,0, n-1) = m*(m+1)/2 - (n-1)*n/2 = (m + n)*(m - n + 1)/2
either m+n or m-n+1 is even, so any number that is the sum of consecutive numbers has to be the product of two numbers, which means that it doesn't hold for any prime number..
So my guess would be it's a prime between 1000 and 2000..
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towr
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Re: Sum of Consecutive numbers
« Reply #3 on: Jan 26th, 2003, 8:01am » |
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hmm.. of course any odd number is the sum of two consecutive numbers :p
(2n+1 = n + n+1)
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Chronos
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Re: Sum of Consecutive numbers
« Reply #4 on: Jan 27th, 2003, 4:30pm » |
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We already covered a very similar question, in the Finger Gauges puzzle. Reasoning from there will suggest at least one answer to the puzzle, and I think it's safe to say that there aren't any more.
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Icarus
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Re: Sum of Consecutive numbers
« Reply #5 on: Jan 27th, 2003, 5:02pm » |
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If n = (2m+1)k, then n= (k-m)+(k-m+1) +...+k+...(k+m).
So any number with an odd factor is the sum of consecutive integers. Pietro has already shown that every sum of consecutive integers has an odd factor.
Hence n is NOT a sum of consecutive integers if and only if n is a power of 2. 1024 is the only power of 2 between 1000 and 2000.
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Chronos
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Re: Sum of Consecutive numbers
« Reply #6 on: Jan 29th, 2003, 11:10am » |
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Except that there might conceivably be other numbers which don't work, if negative integers are not allowed. I don't think that this will happen, but it needs to be considered.
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Icarus
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Re: Sum of Consecutive numbers
« Reply #7 on: Jan 29th, 2003, 8:07pm » |
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Even if negatives are not allowed, the answer is the same: what positive integers are the sums of two or more consecutive natural numbers?
If n is a power of 2, then by Pietro's post it is impossible.
Let n=k(2m+1) with m>0. The expression I gave in my last post is all positive as long as m < k.
Now,
2m+1 =
m + m+1
m-1 + m+2
m-2 + m+3
. . .
m-k+1 + m+k
Summing these gives
n =k(2m+1) = m-k+1 + m-k+2 +...+ m + ... m+k
Which is all positive as long as m > k. So, n is a sum of consecutive positive integers iff it is a positive integer and not a power of 2.
Modified to remove erroneous first line.
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| « Last Edit: Jan 30th, 2003, 5:12pm by Icarus » |
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towr
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Re: Sum of Consecutive numbers
« Reply #8 on: Jan 30th, 2003, 12:04am » |
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2 = -1 + 0 + 1 + 2 thus 2 is the sum of consecutive integers :p
And any other number likewise..
n = sum(i, i, -|n-1|, |n-1|) + n
0 is a special case, since any k!=0 will work in sum(i, i, -|k|, |k|) + 0
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Icarus
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Re: Sum of Consecutive numbers
« Reply #9 on: Jan 30th, 2003, 5:44pm » |
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Good catch towr!
on Jan 26th, 2003, 7:11am, Pietro K.C. wrote:| The sum of N consecutive numbers is odd, if N is even... |
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I read this, checked the obvious case, and accepted it without looking closer until towr posted his counterexample.
The sum of N consecutive numbers is odd if N=2 mod 4
The sum of N consecutive numbers is even if N=0 mod 4.
Obviously Chronos is right: the puzzle requires restriction to non-negative integers. (My apologies for disputing this earlier. I have changed my last post to remove this unfortunate erroneous reply.  )
By my previous post, every number not a power of 2 is the sum of consecutive positive integers. So the only possible answer is 1024. However it has not been shown after all that 1024 is not also the sum of consecutive positive integers.
However: (m+1) + (m+2) + ... + (m+k) = km + k(k+1)/2 = k(2m+k+1)/2.
We are given that k>1.
So if k is odd, it is a factor of the sum. If k is even, (2m+k+1) is an odd factor of the sum provided 2m+k+1 != 1. But if we are restricted to positive integers, m > 0, and so 2m+k+1 > 2. Thus every sum of 2 or more positive integers is not a power of 2.
1024 is the unique answer to the (corrected) riddle.
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| « Last Edit: Jan 30th, 2003, 5:47pm by Icarus » |
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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