What four numbers can equal 22, 20, 32, 30, 23, 25, and 24 when added?

NayNayplaysgame

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So, I don't quite know if I would say that this is algebra but if doesn't seem like arithmetic either so I'm just putting it here because of the variables. Anyway, I have a four by four grid with four variables (A, B, C, and D) scattered about the square with the numbers that i should get by adding the columns and rows corresponding to them and it looks like this:

ccad =22

cdda =20
aadd =32
cbba =30

cacc =24
badc =25
bdda =23
adad =32

(It's the same graph just sideways to show the columns easier.)

So far, I've gotten A=9, B=8, C=5, and D=3. However, that does not work for the 32's, and after around half an hour I still haven't been able to fix that without causing more problems.
Thanks.

 
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I think I'd approach the problem by first bounding in the values. Were you given any bounds on the values to start with? For instance, did the problem say "you may only use the numbers 1-9" or apply similar constraints? Did it say whether two variables can have the same value? For the moment, I'll assume the only restriction that was given is that all values must be positive whole numbers, purely for simplicity's sake. Just to help you get started, my first step was as follows:

One of the columns tells us that a + 3c = 24. Therefore, we know that the maximum value of c is 7. If c were 8 or more, then there's no possible value for a to make the equation work. From there, we can work out constraints on a. Because I've assumed that all values must be positive whole numbers, there are only seven possible values for c, and thus only seven possible values for a.​ Then we can test these possible values by plugging them into the other equations and eventually lock in the only possible values. And once we have set-in-stone values for a and c, that will make finding b and d that much easier.
 
...

1. ccad =22

2. cdda =20
3. aadd =32
4. cbba =30


5. cacc =24
6. badc =25
7. bdda =23
8. adad =32
....

As ksdhart hints at, some rules are needed. Must a-d be integers? positive integers? ...

If they must be integers, since
a = 24-3c
a must be divisible by 3 so one could start there and get your solution, i.e. (a, b, c, d)=(9, 8, 5, 3) starting with a=9 and thus c=5. Or one could start with a=3 and thus c=7 to get (3,10,7,5) but that doesn't work for 32 either.

So another approach. Labeling your equations 1-8 (see the red above) use 5 to get
5: a = 24-3c
4: b = (30 - c - a)/2 = (30 - c - 24 + 3c)/2 = (6 + 2 c)/2 = 3 + c
1: d = 22 - 2 c - a = 22 - 2 c - 24 + 3 c = -2 + c
3: a = (32 - 2 d)/2 = ... uh-oh
 
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