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Hi ganesh,

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25 seconds!

Hello,

Hi ganesh!

Modular arithmetic is something I don't fully understand; so maybe this isn't right:

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**Denominator**- Replies: 1

Apart from doing ganesh's excellent math problems (thank you ganesh!), I have been doing some Project Euler problems.

The problems on this site require programming in order to solve (99% of the time!). For example, the first problem reads:

If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.

Find the sum of all the multiples of 3 or 5 below 1000.

Over the past few years I have worked on their problems and have solved 192 out of their 750+ problems.

But right now I've hit a roadblock, and haven't been able to solve any in weeks. I'm looking for new (or existing) members to add on Project Euler, which would hopefully motivate me to continue working on the problems.

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Hello!

Hello,

Hi ganesh,

Hello,

Hello ganesh,

Hello, here are my thoughts:

Sounds like you are only looking at possible amounts of 1s, 2s and 3s (as opposed to a specific order of slots).

^ expanding that what I mean is that it appears you aren't differentiating between [ten 1s first then ten 3s], and [ten 3s first then ten 1s].

Looking only at a sum total of 40, here's one way to find the number of possible collections:

Start off with twenty slots of 2 (sum is 40). Then convert two 2s into one 1 and one 3 (so as to conserve the sum, as 2+2 = 1+3). This is another possible collection of numbers. You can repeat this process nine more times. Therefore there are 11 slot collections that sum to 40.

Let (x, y, z) be the numbers of 1s, 2s and 3s respectively. Then the collections are {(0,20,0), (1,18,1), ... (10,0,10)}, which give 11 possible ways to sum to 40.

Your question asks about combinations that don't exceed 40, so this is not the full answer. But hopefully this has given you a start or possibly new ideas on how to answer your question

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