# Find the inverse ratio of a group of numbers.

(I'm very open to suggestions to a better title to my question)

I have four numbers, 10, 20, 30 and 40. I need to multiply each of then by some number so that they are all equal. However the sum of the numbers I multiply by has to be equal to 1.000. As shown in the following table.

$$\begin{array}{r | r r} n & \text{Multiplier} & \text{Total} \\ \hline 10 & ? \\ 20 & ? \\ 30 & ? \\ 40 & ? \\ \hline 100 & 1000 \end{array}$$

The way I did this was to divide each 1 by n and then sum all of those to get the ratio. Then divide the ratio by the total ratio. (I hope I'm still making sense). As in the following table.

$$\begin{array}{r | r r} n & \text{Multiplier} & \text{1/n} & \text{Ratio} & \text{Total} \\ \hline 10 & ? & 0.100 & 0.48 \\ 20 & ? & 0.050 & 0.24 \\ 30 & ? & 0.033 & 0.16 \\ 40 & ? & 0.025 & 0.12 \\ \hline 100 & 1000 & 0.208 \end{array}$$

When I have this it's quite straightforward of course to just use the ratios and multiply by 1000 and everything works out great.

My question is however, is it possible to go straight from the information in the first table and find out the ratio or the number required to multiply by without doing 1/n for each number first?

Edit: To clarify what the final table should look like

$$\begin{array}{r | r r} n & \text{1/n} & \text{Ratio} & \text{Multiply n with} & \text{Total} \\ \hline 10 & 0.100 & 0.48 & 480 & 480\\ 20 & 0.050 & 0.24 & 240 & 480 \\ 30 & 0.033 & 0.16 & 160 & 480 \\ 40 & 0.025 & 0.12 & 120 & 480\\ \hline 100 & & 0.208 & 1000 \end{array}$$

• The l.c.m of the given numbers is $120$. So one needs to multiply by $12, 6, 4, 3$ to $10, 20, 30, 40$ respectively to get $120$ each, and $12+6+4+3=25$. Now multiply by $12\times 40, 6\times 40, 4\times 40, 3\times 40$, and sum of these numbers is $1000$. – Krish Oct 19 '17 at 10:12
• Thanks, but I don't think that's what I'm after though. I'm looking to multiply 10 with n1, 20 with n2, 30 with n3 and 40 with n4, and they should all be equal and all those n should add up to 1000. – Eyeslandic Oct 19 '17 at 12:20
• Take $n_1 = 12 \times 40, n_2= 6 \times 40, \cdots$; and $n_1+n_2+n_3+n_4=1000$. Also $10\times n_1 = 20 \times n_2 = \cdots$. – Krish Oct 19 '17 at 12:23
• @Krish Thanks, I'm closer to understanding it now! – Eyeslandic Oct 19 '17 at 12:27