Question

Solve the following cryptarithm and find the values of A and B.
\[\begin{align}
  & +\left| \!{\underline {\,
  \begin{matrix}
   3 & 7 \\
   A & B \\
\end{matrix} \,}} \right. \\
 & \,\,\,\,\begin{matrix}
 \,\, 9 & A \\
\end{matrix} \\
\end{align}\]

Answer 1

Hint: For this question, we have two cases. In the first case, we have \[7+B\le 9\]. We have two equations \[7+B=A\] and \[3+A=9\] . We have to find the values of A and B and now, we have two equations. Using these two equations, find the values of A and B. In the second case, we have \[7+B>9\]. We have two equations \[7+B=A\] and \[1+3+A=9\] . We have to find the values of A and B and now, we have two equations. Using these two equations, find the values of A and B.

Complete step-by-step solution-
Cryptarithm is a type of mathematical puzzle in which the digits are replaced by some symbols or letters of the alphabet. Usually, each letter or symbol is replaced by a unique digit, and also each letter has a different value from the other letters.
According to the question, we have a summation of two numbers.
The summation of one’s digit can either be less than equal to 9 or greater than 9.
Let us continue with the first case which means \[7+B\] is less than equal to 9.
\[7+B=A\] ……………..(1)
\[3+A=9\] ………….(2)
From equation (2), we can find the value of A. Moving the constant term 3 to the RHS of the equation (2), we get
\[\begin{align}
  & 3+A=9 \\
 & \Rightarrow A=9-3 \\
 & \Rightarrow A=6 \\
\end{align}\]
Now, we have \[A=6\] .
\[\begin{align}
  & +\left| \!{\underline {\,
  \begin{matrix}
   3 & 7 \\
   6 & B \\
\end{matrix} \,}} \right. \\
 & \,\,\,\,\begin{matrix}
   9 & A \\
\end{matrix} \\
\end{align}\]
Putting \[A=6\] in equation (1), we get \[7+B=6\] .
In the first case, we have already assumed that \[7+B\] is less than equal to 9.
We can see that the unit digit of \[7+B\] can be equal to 6 if B is either equal to 9 or -1. But B cannot be negative. So, B = -1 is not possible.
Taking B = 9, we get
\[7+B=9+6>9\]
We got \[7+B\] greater than 9.
But in the first case, we have assumed that \[7+B\] is less than equal to 9 but here we got \[7+B\] greater than 9 which is a contraction.
Therefore, the first case is not possible.
Let us continue with the second case which means \[7+B\] is greater than 9. If \[7+B\] is greater than 9 then we have to add 1 as carry on ten’s place
\[\begin{align}
& +\left| \!{\underline {\,
\begin{matrix}
1 & \\
3 & 7 \\
A & B \\
\end{matrix} \,}} \right. \\
& \,\,\,\,\begin{matrix}
9 & A \\
\end{matrix} \\
\end{align}\]
\[7+B>9\] ……………..(3)
\[1+3+A=9\] ………….(4)
\[7+B=A\] ……………..(5)
From equation (4), we can find the value of A. Moving the constant term 3 and 1 to the RHS of the equation (4), we get
\[\begin{align}
  & 1+3+A=9 \\
 & \Rightarrow A=9-4 \\
 & \Rightarrow A=5 \\
\end{align}\]
Now, we have \[A=5\] .
Putting \[A=5\] in equation (5), we get
\[7+B=5\]
We can see that the unit digit of \[7+B\] can be equal to 5 if B is either equal to 8 or -2. But B cannot be negative. So, B = -2 is not possible.
Taking B = 8, we get
\[7+B=7+8>9\]
It also satisfies our assumption that \[7+B\] is greater than 9.
So, B = 8 is possible.
Hence, the value of A is 5 and the value of B is 8.

Note: In the second case of this question, one can make a mistake by not adding 1 at the ten’s digit which is wrong. If the summation of one’s digit is greater than 9 then we add 1 as carrying to the ten’s digit. So, in case 2nd our equation will be \[1+3+A=9\] not \[3+A=9\] .
Cryptarithms are puzzles in which the digits are replaced by letters or symbols in an arithmetical calculation. Algebraic expressions might be considered as a type of cryptarithms, but algebra is not generally considered to be mathematically recreational.