# 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, \(B + F + K = 27\).