Numbers

Happy girl

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Find a number between 20 and 50 which cannot be made by summing consecutive whole numbers ??
 

mmm4444bot

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Hello. What class are you taking? Please share what you've already tried or thought about. Thank you.

 

Otis

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Hi HG. There are 29 Whole numbers between 20 and 50. You don't need to check them all; a lot of them can be eliminated by considering patterns. Have you experimented?

If not, then here's something easy you could try. Take any pair of consecutive Whole numbers and add them. Now pick a different pair of consecutive Whole numbers and add them. Continue doing this until you recognize a pattern. Can you see how the pattern shows that more than half of the possibilities in your exercise can be eliminated?

There are other patterns that can eliminate additional possibilities. Compare the average of three consecutive Whole numbers to their total. Do you see a relationship? Experiment with beginning sets like {1,2,3}, {2,3,4}, {3,4,5}, {4,5,6}, etc.

Here are some additional questions to consider. Starting with 1, how many consecutive Whole numbers do you need to add to get a total greater than 20? What happens if you increase that total by adding the next Whole number? Is the new total still below 50? What happens if you start with 2 instead of 1? What happens if you start with 3?

Organize your experimental work so that you can see prior calculations. That can help to reduce the amount of arithmetic needed for eliminating all remaining possibilities but the answer.

PS: You've posted on the Arithmetic board. Let us know if you're allowed to use algebra. There's a nice pattern for checking possibilities quickly that arises from basic symbolic reasoning.

😎
 

HallsofIvy

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"Two consecutive whole numbers" must be of the form n and n+1. Their sum is 2n+ 1 which is necessarily an ODD number.

Any EVEN number "cannot be made by summing two consecutifve whole numbers".

Any even number between 20 and 50 will satisfy this.
 

Otis

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Any EVEN number "cannot be made by summing two [consecutive] whole numbers"
Hi Halls. The op doesn't contain the word 'two'.

I worked out a nice progression (if using algebra), but I'm going to wait a few more days before posting it.

:)
 

Jomo

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I liked this problem so I looked into it a little.

It is easily seen that n+(n+1)+(n+2)+...+(n+k) = (2nk+k^2+2n+k)/2

Let's see if such a sum can ever be 32.

(2nk+k^2+2n+k)/2 = 32 iff (2nk+k^2+2n+k) = 64

Solving for n we get n=(64-k^2-k)/(2(k+1)) = 32/(k+1) - k/2.

If k is even, then k/2 is an integer. So n is an integer iff 32/(k+1) is an integer. Since 32 = 2^5 and k+1 is odd, 32/(k+1) is not an integer, concluding k is not even.

Assuming k is odd, it is clear that k+1 is even. Let k=2r+1 for some integer r.

n = 32/(2r+2) - (2r+1)/2 = (31-2r^2 -3r)/[2(r+1)]. The remainder from this division is -16/(r+1).

We actually need -16/(r+1) to be an integer. So r is in {1,3,7,15}

Then k is in {3, 7, 15, 31}.

A quick check shows that none of these k values results in n being an integer. We conclude that 32 is not obtainable by a sum of consecutive integers.
 

Dr.Peterson

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I used a spreadsheet to find the answer (not just for numbers from 20 to 50, but starting at 1 and going some distance) by more or less brute force, and it appears that all numbers that can't be formed as a sum of consecutive numbers form a very simple sequence:
powers of 2 -- that is, numbers with no odd prime factors; 32 is the only one between 20 and 50
When I saw this, I realized I've see that fact before, and maybe even figured it out myself at that time (not by brute force).

I can come up with a proof, but it's not as elegant as I'd like, though it's a lot cleaner than Jomo's.

Now that I think about it, I suspect it arose from trying to find a method to find such a summation for any given number. For instance, given the number 35, find a sum of consecutive integers that has that sum. Since 35 = 5*7, one possibility is the sum of 5 numbers whose average is 7, namely 5+6+7+8+9, and another is the sum of 7 numbers whose average is 5, namely 2+3+4+5+6+7+8. On the other hand, for 22 = 2*11, we could have 2 numbers whose average is 11 (which wouldn't be integers), or 11 numbers whose average is 2 (which would be -3+-2+-1+0+1+2+3+4+5+6+7); but the fact that those don't work doesn't mean there is no such sum, because we can reduce this to 4+5+6+7.

When can't you do it? How do you prove this simply? (Again, what I've done here doesn't lead yet to a conjecture we can try to prove, or to an immediate proof. But I think it did when I worked through this years ago.) I'm probably right on the edge, but I have to go now!
 

Otis

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Another puzzle: Which number between 20 and 50 can be expressed as a sum of consecutive Integers in the greatest number of ways?

:sneaky:
 

Cubist

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It is easily seen that n+(n+1)+(n+2)+...+(n+k) = (2nk+k^2+2n+k)/2
= (1 + k) (2n + k) / 2

This can never be ( 2*2*2*...*2 ) ( 2*2*2*2*...*2 ) / 2

To prove this, let (1+k) = 2^z with z≥1. Then (2n + k) = (2n + 2^z - 1) = an odd number
 
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