Forgive me if this question is very simple or if my question is in wrong format.
I Have a set of random numbers let's say 7, 9, 1 And this could go on but only positive integers.
I have to add these numbers in a pyramid form...so that 7+9 =16, 9+1 =10 and so on..
Every time I move up a layer in the pyramid I have to remove the numbers in the layer below.
Once I am left with only two numbers to add at the top of the pyramid, I stop adding.
Now, I need to make my way down the pyramid again and figure the exact numbers I used to add up.
Can someone help me solve for this.
My idea is, If I am left with two numbers at the top of the pyramid I am left with 16 and 10... I know that $$ 1 6 =A + B \quad \text{and} \quad 10 = B + C. $$
I need to solve for A, B and C.
I am happy to clarify if nor well understood.
Thanks in advamce
Looks like it is impossible. The system of these equations has multiple solutions. You can take $B=2$ => $A=14, C=8$ - that's one solution. Or you can take $B=3$ => $A=13, C=7$ -it's a different solution.
As a rule of thumb - number of (independent) equations should be equal to number of unknown variables - in this case there is a single solution. If number of equations is larger - there is no solutions. If it is smaller - there are multiple solutions. This is not a strict rule (the equations could be dependent) but usually it works.
Things are much more complicated if there is a requirement that all variables are integers, but simple counterexample works in this case: you can't solve the system of equations.