Puzzles

106

196

495

216

Adding up all three sums of pairs (284) will give twice the sum of the three numbers. So the sum of the three numbers is 142.

123

The two numbers between 14 are a cube number and a triangle number: these must be 8 and 6. Next you can see that 23 is the sum of 8, two times a triangle number and a square number: the triangle and square numbers must be 3 and 9.

Next, call the square number at the bottom left \(a\), the square number at the top left \(b\), and the triangle number at the top right \(c\). Adding upwards, we find that \(a+b+45=106\) and \(a+141+c=198\); and so \(a+b=61\) and \(a+c=57\). The only two square numbers that add to 61 are 25 and 36. Therefore \(c\) must be 21 or 32, but must be 21 as 32 is not a triangle number. And so \(a\) is 36 and \(b\) is 25.

Putting all these numbers into the tree gives the top number as 433. This is fitting because 433 is in fact a star number:

Both the units and tens columns contain \(Y+Z\). The results are different (\(X\) and \(Z\)), so \(Y+Z=X+10\) and \(Z=X+1\) (because the 1 carries into the next column).

Therefore, \(Y+X+1 = X+10\), so \(Y=9\).

From the hundreds column, we see that \(X+X+1=Y\), so \(X=4\) and \(Z=5\).

Which digits \(X\), \(Y\) and \(Z\) fill this sum?