Juliana wants to assign each of the 26 letters $A, B, C, D, \ldots, W, X, Y, Z$ of the alphabet a different non-zero numerical value, such that $A \times C=B, B \times D=C, C \times E=D$, and so on, up to $X \times Z=Y$.

a) If Juliana assigns the values 5 and 7 to $A$ and $B$, respectively, what will be the values of $C, D$, and $E$?

b) Show that $G=A$, regardless of the values Juliana assigns to $A$ and $B$.

c) If Juliana assigns values to $A$ and $B$ such that $A \times B=2010$, what will be the value of the product $A \times B \times C \times D \times \cdots \times W \times X \times Y \times Z$?
Solution

a) Substituting $A=5$ and $B=7$ into $A \times C=B$, we get $5 \times C=7$ and it follows that $C=\frac{7}{5}$. We can now find $D$ by substituting the values of $B$ and $D$ into $B \times D=C$; we obtain $7 \times D=\frac{7}{5}$ and thus $D=\frac{1}{5}$. Finally, from $C \times E=D$ we have $\frac{7}{5} \times E=\frac{1}{5}$ and see that $E=\frac{1}{7}$.

b) Multiplying the expressions $A \times C=B$ and $B \times D=C$ we get $A \times B \times C \times D=B \times C$; since $B$ and $C$ are different from 0, we conclude that $A \times D=1$, or $D=\frac{1}{A}$. Similarly, multiplying the expressions $B \times D=C$ and $C \times E=D$ we get $B \times E=1$, or $E=\frac{1}{B}$. Repeating this reasoning, we see that each letter starting from $D$ is the reciprocal of the letter that appears three positions before it; in particular, $G=\frac{1}{D}=\frac{1}{\frac{1}{A}}=A$.

c) The previous item shows us that $C \times D \times E \times F \times G \times H=C \times D \times E \times \frac{1}{C} \times \frac{1}{D} \times \frac{1}{E}=1$; the same reasoning shows that the product of any six consecutive letters is equal to 1. We then have:

$A \times B \times C \ldots \times Y \times Z=A \times B \times(C \times \ldots \times H) \times(I \times \ldots \times N) \times(O \times \ldots \times T) \times(U \times \ldots \times Z)=A \times B=2010$

since all the products in parentheses are products of six consecutive letters, so they are all equal to 1.

Note: We notice that this problem depends on the fact that, once the values of $A$ and $B$ are fixed, the sequence of the values of the letters of the alphabet is $A, B, \frac{B}{A}, \frac{1}{A}, \frac{1}{B}, \frac{A}{B}, A, B, \frac{B}{A}, \frac{1}{A}, \frac{1}{B}, \frac{A}{B}, A, B, \ldots$