# How to split a number up by ratios

I've looked everywhere for a solution, but the problem is I'm not a mathematician so have no frame of reference.

I'm trying to split a number up by ratio. For example, if you take the number 10:

• If the smallest partition is 3, the result is 3, 7
• If the smallest partition is 2, the result is 2, 3, 5
• If the smallest partition is 1, the result is 1, 2, 3, 4

Any help would be much appreciated.

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I don't know what you mean by "split up a number by ratio." I don't understand why, if the smallest part is 2, the result isn't 2,4,4 or 2,8. I don't understand why, if the smallest part is 1, the result isn't 1,4,5, or 1,3,3,3, or 1,9, or 1,2,7, or 1,2,2,2,3, or something else. What are you really trying to say? –  Gerry Myerson May 29 '12 at 12:26
I see what you mean, the first sequence confuses things. I'll change it to 3, 7, i.e., the next number has to be bigger each time. Does that make things clearer? –  habitullence May 29 '12 at 12:43
I still don't quite see it. You are given a base number and a 'smallest partition,' and you want to write the base number as a sum. The summands start with the 'smallest partition' and must get bigger at each step. That's what I see so far. But I don't know what your question is. And this isn't well-defined. For $13$, for example, with 'smallest partition $1$, I have both $1, 3, 4, 5$ and $1, 2, 4, 6$ I suppose what I want to say is: what's your question? And what's the source of the question? –  mixedmath May 29 '12 at 13:00
Thanks @mixedmath, didn't think of that. I guess the second example would be better cause the increase is more evenly distributed. I'm work on creating ranks, e.g., in a computer game. –  habitullence May 29 '12 at 13:13
So let's say I have 100 ranks in a system and a total of 1,000,000 XP. I want to create a series of ranks that are increasingly harder to obtain and gaining 1,000,000 XP means you're at the highest rank. Also the lowest rank threshold can be set to a number, e.g., 500. Does that make things clearer? –  habitullence May 29 '12 at 13:19

## 1 Answer

From the comments, you want 100 numbers satisfying $$500=m_1\lt m_2\lt\cdots\lt m_{100}$$ and $$m_1+m_2+\cdots+m_{100}=1000000$$ and you want the numbers $m_{i+1}-m_i$ to be roughly equal. If they were exactly equal, say, $m_{i+1}-m_i=d$ for all $i$, you'd have a 100 term arithmetic progression, starting with 500, and summing to 1000000. You'd have $m_{100}=500+99d$, and you'd have $$50(1000+99d)=1000000$$ That's $1000+99d=20000$, $99d=19000$, $191\lt d\lt192$ if I've done my arithmetic right. So you could make some of the steps 191, and the rest of them 192.

Now, you might want round numbers. In that case, you could make some of the steps 150, some of them 200 or 250. You've got a fair bit of room to play around with the numbers. Try out some schemes, and see what looks nice to you.

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Thanks, I understand what you're saying up to, "You'd have m100 = 500 + 99d, and you'd have...". I'm try to get something like: - Rank 1, 500 XP - Rank 2, 1500 XP - Rank 3, 3000 XP - Rank 100, 1,000,000 XP –  habitullence May 29 '12 at 13:53
Not sure how you got from m100=500+99d to 50(1000+99d)=1000000 –  habitullence May 29 '12 at 14:02
OK, I've been learning. Understand now. –  habitullence May 29 '12 at 14:54