What digit in place of $\red{B}$ would make this sum true?

Looking at the last column, we have that $\red{B}$ $+ 6$ ends in a $1,$ so we must have $\red{B}$ $= 5$. Checking the rest of this cryptogram, we verify that $15 + 56 = 71$ is true.

Hence, $\red{B}$ $= 5:$

$\Large \begin{array} { c c c } & 1 & \red{5} \\ + & \red{5} & 6 \\ \hline & 7 & 1 \\ \end{array}$

Not all cryptograms unravel in a single step like the one above. For example, consider divisibility rules — what numbers can multiply together to make a specific product?

$\Large \begin{array} { c c c } & \green{B} & \pink{A} \\ \times & & 6 \\ \hline 1& 6 & \pink{A} \\ \end{array}$

Since we know that the product is a multiple of $6,$ it must be both even and divisible by $3.$ Therefore, the only final digits possible are $2$ and $8:$ we can't have an odd final digit because then the number wouldn't be even, and we can't have $0, 4,$ or $6$ because then we'd have a number that's not divisible by $3.$

So we have two possibilities: $\green{B} 2 \times 6 = 162$ or $\green{B} 8 \times 6 = 168.$ However, $162 = 6 \times 27$, so the first option won't work. What about the second? Well, $168 = 6 \times 28$, so that must be the answer: $A = 8$ and $B=2$.

Here are a couple more advanced techniques to consider:

• Equations: Convert the problem into equations that take the place value of the letters into account. For example, $R2D2 = 1000R + 200 + 10D + 2.$

• Carry digits: Be aware of how carry digits work — when adding two numbers, you carry the ‘overflow’ from one place value to the next if the sum is greater than or equal to $10.$

• Check cases: Organize and eliminate possibilities — keeping track of the possibilities carefully and in an organized way!

Now try your hand at the cryptogram below.