The Top Brick

Let the numbers in the bottom row be $3, a, b, 2$ and work toward the top as shown in the image.

The top brick is $3+3a+3b+2=3(a+b)+5$ .

But we know that the center brick in second row from the bottom is $a+b=32$ .

So the answer is $3\cdot32+5=\boxed{101}$ .

Call the two bottom numbers $x$ and $32-x$ . Then perform the correct additions and the $x$ s will cancel out, leaving the answer as $\boxed{101}$ , as shown below

Fourth row: The two blanks add up to $32$ .

Third row: The blanks left and right add up to $5 + 32 = 37$ .

Second row: The blanks from the third row contribute 37, and the middle value of 32 contributes twice. This makes the total $37 + 2\times 32 = 101$ .

First row: The top block is the total of the second row, i.e. $\boxed{101}$ .

If there is a unique solution (and there must be), then it doesn't matter what numbers you put in the bottom row so long as they add to 32. In fact picking any two numbers and quickly adding up the pyramid may be faster than developing and solving the problem using the general algebraic approach shown in other solutions.

No matter what number you assign for the bottom 2 bricks, the final sum on for the block at the top is constant. This is because if you add +1 to one brick, you'll then have to subtract -1 away from the other one and therefore the overall effect is 0. So, we can just pick 2 random numbers and assign them for the 2 bricks and work it out all the way to the top as long as the 2 numbers add up to 32.

Just put the appropriate numbers

3 + 2 + 32 * 3 = 101.

The 3 and 2 are on the edge so they are carried up to the top as they are.

The 32 is used 3 times, once it is split and part of it makes up each brick to the left and right of it. It is then duplicated twice for the two bricks above it.

For these kinds of puzzles they can be solved by working out how many times the value for each brick is used, of course it's much harder to work out on larger puzzles.

The top brick is $3+3(a+b)+2=3+3 \times 32 +2=3 \times 33+2=99+2=101$

This is actually a very simple puzzle, since you can insert any two numbers (a and b for example) to the bottom row, as long as a + b = 32.

Bottom row: 3, 32-x, x, 2 Second row: 35-x, 32, x+2 Third row: 67-x, 34+x Top row: 101

$a = b + c$

$b = d + 32, c = 32 + f$

$d = 3 + h, 32 = h + i, f = i + 2$

$a = 3 + h + 32 + 32 + i + 2 \Rightarrow (3 * 32) + 5 = 101$

In the bottom line suppose the numbers are =a,b Here we get, a+b=32 So the number on the to would be =3+a+32+2+b+32 =a+b+69 =32+69 =101

Let the blanks in the bottom row be $l$ and $r$ for left and right, respectively.

The square to the bottom-left of the top is the sum of the ones below itself, namely $(3+l)$ and 32, and the square below the top to the right is the sum of 32 and $(r+2)$ . Add and simplify to find that the top square $= l + r + 69$ .

The square marked $32$ is the sum of the two below it, so $l + r = 32$ .

We now have two equations. Let me add parentheses to give a hint: $top = (l + r) + 69$ $(l + r) = 32$

Substitute the quantity $(l + r)$ into the first equation from the second: $top = (l + r) + 69 = 32 + 69 = 101$ .

I just kept splitting 32 in different ways until I found a way that worked.

Just in case somebody would like to feel it instead of solving it in algebraically, they should keep in mind that the first and foremost part of breaking the number 32 into two components would not depend on how you break it. It is because all numbers are being added in every successive step, so there is no loss of any existing unit in the process, and the answer comes out the same no matter how you break 32.

Pick any 2 numbers that add up to 32 and place them in either box below 32 and simply solve up .

I figured that in order for this to have a solution, it wouldn't matter what the two missing numbers were as long as they added to 32. So tried 32 and 0 and then tried 16 and 16 and low and behold, 101...