How create a database with the sums of all possible combinations of 2 to 8 elements from 28 different numbers?

Question

I have 2 columns with 28 different numbers each, I need to create a database using excel program with the sums of all possible combinations between 2 and 8 elements from 28 numbers in each column and compare the results to find identical results.
For example in the first column the sum of 3550+50+500+125+18.40=4234.40 its identical with the sum of 2000+2220+14.4=4234.4 in the second column. In other words I need to find identical sums with different number of elements between both columns.
I don´'t know how create a program in excel that let generate automatically all possible sum of combinations from 2 to 8 elements from 28 quantities in both columns and then compare one with other in order to find identical results.
Can someone help me?
Thank you.

Answer 1

With out programming this is gonna be brutal. Hope you can do cutting and pasting and don't mind dragging a lot.....

You can use a feature in excel which sequentially inserts numbers by dragging called AUTOFILL.

Start in cell A1 and type 1
goto the cell below it A2 and type a 2

Left click on cell A1 and drag the mouse to highlight cells A1 & A2
Release the mouse
Cells A1 & A2 are now surrounded by a bold box with a small square at the lower right corner

Using the mouse click and hold the small square and drag the mouse downward. As you do the cells will be surrounded by a light grey box and a yellow highlighted counter will appear next to the mouse. When you release the mouse button Excell will load the previously empty cells with sequenced numbers

The key is showing Excel the pattern you want with the first group of number you enter. So if you want even numbers do the same as above by enter the number 2 in cell A1 and 4 in Cell A2
Now Excell will sequence by 2's

Use this technique to load numbers 1 through 28
Copy these numbers and paste them into the next 28 cells
Do this 28 Times

You will now have a column of numbers 1 to 28 repeating 28 times.

Highlight these cells now grab the little square and drag over to the right one cell

You will now have a copy of column A in Column B

Now you need to sort Column A so that the number 1 is with the first group of 28 in column B ect...

Click on the letter B to high light the entire column. Then in the menu
Click DATA then SORT. On the popup select use current selection click OK
Then Click OK again.

Column A will now be sorted grouping the 8 copies of each number together.


Now insert a new column at Column A

Copy the data in columns B & C 28 times
Copy col B in to col A
Sort A

Do this over and over util you have your 728 rows by 8 Columns


You now have your repeating pattern of 28 numbers
Use find and replace later to change the values 1 to 28 with what ever other number you need.

Answer 2

enable P(a, b) = ? x for all a < x < b Lemma a million: with techniques from the easy guidelines of Set concept, for any constructive sequence a < b < c, P(a, c) = P(a, b) * b * P(b, c). All of my arguments follow from this assumption. ----------- (2) ----------- P(0, 2) = P(0, a million/2) * a million/2 * P(a million/2, a million) * a million * P(a million, 2) for each x, a million/2 < x < a million, there exists precisely one counter-section a million/x, such that a million < a million/x < 2. on the grounds that that is a bijection, we may be able to exhibit P(a million/2, a million) * P(a million, 2) as: ? (a million < x < 2) x * (a million/x) = ? (a million < x < 2) (a million) = a million with techniques from substitution, P(0, 2) = a million/2 * P(0, a million/2) P(0, a million/2) techniques 0, so P(0, 2) techniques 0 I justify P(0, a million/2) = 0 with techniques from the actual shown truth that P(0, a million/2) is an unlimited made from factors all <= a million/2. So, for any constructive integer n, P(0, a million/2) <= (a million/2)^n. (a million/2)^n -> 0 as n -> infiinity, so P(0, a million/2) -> 0 ---------- (3) ---------- i'm puzzled as to why you imagine this example is mistaken - what's incorrect with it? The "Riemann" attitude works: enable f(0) = a million, f(x) = x * f(x - a million/n), n -> infinity f(ok/n) = ok! / ok^n, f(2) = (2n)! / (2n^2n) -> 0 as n -> infinity also, how can we exhibit this actual countless product as a sum? not confident a thanks to attempt this. @Scythian: you've hit the mark - the finished complicated crux of this example. the element is, you could map any period (a, b) onto (c, d) utilising the bijection f(x) = c + (b-x)*((c-d)/(b-a)). in truth, we may be able to map any finite period (a, b) onto R = (-infinity, +infinity) with techniques from the bijection: f(x) = 0 even as x = (a + b) / 2, a million / (2x - a - b) otherwise Does this make the issue unsolvable? i imagine not - i imagine it only skill we'd want to be careful how we interpret the "countability" of countless gadgets. that is organic for us human beings, residing in a Cartesian, Euclidean international, to imagine that linearity is a valid and inevitably acceptable mapping for any set of numbers. although, this isn't the case, as is shown interior out with techniques from rigorous learn of Set concept. i am going to't fake to be an authority in this container, yet i have heard adequate from experienced mathematicians to understand the basics.