# Problem of the Week Problem D and Solution Blocked Numbers

## Problem

Twelve blocks are arranged as illustrated in the following diagram.

Each letter shown on the front of a block represents a number. The sum of the numbers on any four consecutive blocks is $$25$$. Determine the value of $$B + F + K$$.

## Solution

Since the sum of the numbers on any four consecutive blocks is the same, looking at the first five blocks, we have $4+B+C+D=B+C+D+E$ Subtracting $$B$$, $$C$$, and $$D$$ from both sides gives $$E=4$$. Similarly, looking at the fifth through ninth blocks, we can show $$J=4$$.

Again, since the sum of the numbers on any four consecutive blocks is the same, looking at the third through seventh blocks, we have $C+D+E+F=D+E+F+5$ Subtracting $$D$$, $$E$$, and $$F$$ from both sides gives $$C=5$$. Similarly, looking at the seventh through eleventh blocks, we can show $$L=5$$.

Once more, since the sum of the numbers on any four consecutive blocks is the same, looking at the eighth through twelfth blocks, we have $H+J+K+L=J+K+L+7$ Subtracting $$J$$, $$K$$, and $$L$$ from both sides, gives $$H=7$$. Similarly, looking at the fourth through eighth blocks, we can show $$D=7$$.

Filling in the above information, the blocks now look like:

We will present two different solutions from this point.

Solution 1:
Since the sum of any four consecutive numbers is 25, using the first 4 blocks \begin{aligned} 4+B+5+7&=25\\ B+16&=25\\ B&=9\end{aligned} Similarly, we can show $$F=9$$ and $$K=9$$.

Therefore, $$B + F + K=27$$.

Solution 2:

We note that the twelve blocks are three sets of four consecutive blocks. Each of these three sets have a total of 25, so the total sum of the blocks is $$3 \times 25 = 75$$.

The sum is also $4 + B + 5 + 7 + 4+ F + 5 + 7 + 4 + K + 5 + 7 = 48 + B + F + K$

This means $48 + B + F + K=75$ or $B+ F + K = 27$ Therefore, $$B + F + K = 27$$.