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**rkonweb****Member**- Registered: 2010-06-24
- Posts: 1

1+3+5+7+9+11+13+15+17+19=100

pick any five digit and by adding them total should be 50.

u can use one digit at once.....

thanks in advance

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,765

Hi rkonweb;

Unless you do something other than just pick from {1,3,5,7,9,11,13,15,17,19} the problem has no solution. 5 odd numbers added together equal an odd number and 50 is not odd.

One answer I found is:

1 + 4 + 9 + 17 + 19 = 50 but it has a 4 which is not in {1,3,5,7,9,11,13,15,17,19}

There are others.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**aawood****Member**- Registered: 2010-06-25
- Posts: 2

As you say, adding any 5 odd numbers would make it impossible, but he mentioned any 5 **digits** rather than **numbers**. We can therefore use that to get 4 odd numbers using 5 digits from the sequence that add up to 50. One I found was 1 + 5 + 9 + 35.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,765

H aawood;

Yes, there are other ways if we begin to interpret the question. How do we know that is what he wants. So far, 3 different answers, all of them correct. Maybe, he wants something else.

Based on his first statement:

1+3+5+7+9+11+13+15+17+19=100. It looks like he figured since 10 odd numbers can sum to 100 and since 50 is 1/2 of a hundred so maybe 5 odd numbers can sum to 50?

Welcome to the forum!

Just between you and me you might have what he wants. But look at this, supposedly a mensa quiz question answer for this problem.

Which is not quite correct, should read:

If you google for this you will find many other strange answers.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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