a+b+c=c+d+e=e+g+h=h+i+j=13? there are unique number for each between 0-9. find e?
A+b+c=c+d+e=e+g+h=h+i+j=13 ?
there are only nine pronumerals in a,b,c,d,e,g,h,i,j, so I must assume that you mean 1-9 inclusive or otherwise 0-8 inclusive.
I'll take 1-9 inclusive first.
If they are unique numbers then
a+ b + c + d + e + g + h + i + j = 1 + ... + 8 + 9
= 45
Adding the four equations produces
a+ b + c + d + e + g + h + i + j + (c + e + h) = 4* 13 = 52
i.e. (c + e + h) = 52 - 45 = 7
Also, (c + d + e) = 13 so therefore
d - h = 6, that is {d, h} = {7, 1}, {8, 2}, {9, 3}
Similarly, (e + g + h) = 13 so
g - c = 6, that is {g, c} = {7, 1}, {8, 2}, {9, 3}
Because the equations are symmetrical, we let c %26gt; h, because c and h don't matter (xP)
i.e. answers for {d, h} = {7, 1}, {8, 2}
answers for {c, g} = {2, 8}, {3, 9}.
so, testing these values and noting values are unique (just look at them, didn't take long...) we can see that the only possible combination that works is {c, g} = {2, 8} and {d, h} = {7, 1} i.e. e = 4
OR
In hindsight, it's a lot easier just to write out the possible combinations of numbers that add up to 13,
i.e. {1, 3, 9}, {1, 4, 8}, {1, 5, 7}, {2, 3, 8}, {2, 4, 7}, {2, 5, 6}, {3, 4, 6}
Only one of these contains a 9, so {1, 3, 9} must be included in the set of equations. This leaves {1, 4, 8}, {1, 5, 7}, {2, 3, 8}, {2, 4, 7}, {2, 5, 6}, and {3, 4, 6}. Looking at {1, 3, 9}, we know that it must be an equation at the furthest left or right, because otherwise {3, 4, 6} and {1, 5, 7} or {2, 3, 8} and {1, 5, 7} will be adjacent to it and the fourth equation won't work. This is easily worked out in less than a minute.
So: {1, 3, 9} is an end point. This means that either 1 or 3 is doubled up in the inner equations. Testing and stepping through valid combinations shows that 1 is doubled up.
{3, 9, 1}, {1, 8, 4} {4, 7, 2} {2, 5, 6}.
i.e. e = 4
Tracing through these combinations, I just wrote them on a piece of paper and used my fingers to follow a given order... not so hard =P
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