Ex.16.1 Q5 Playing with Numbers Solutions - NCERT Maths Class 8

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Find the values of the letters in the following and give reasons for the steps involved.

\[\begin{align}{A \;{B}} \\ { \times \;\;{3}} \\ \hline C\, A\, {B} \\ \hline\end{align}\]

 Video Solution
Playing With Numbers
Ex 16.1 | Question 5

Text Solution

What is known?

Multiplication operation of two numbers

What is unknown?

Value of alphabets i.e. \(A, B\) and \(C.\)


Each letter in the puzzle must stand for just one digit. Each digit must be represented by just one letter.


The multiplication of \(3\) and \(B\) gives a number whose one’s digit is \(B\) again.

Hence, \(B\) must be \(0\) or \(5.\)

Let \(B\) be \(5.\)

Multiplication of first step \(= 3 × 5 = 15\)

\(1\) will be a carry for the next step.

We have, \(3 × A + 1 = CA\)

This is not possible for any value of \(A\).

Hence, \(B\) must be \(0\) only. If \(B = 0\), then there will be no carry for the next step.

We should obtain, \(3 \times A = CA\)

That is, the one’s digit of \(3 \times A\) should be \(A\). This is possible when \(A = 5\) or \(0.\)

However, \(A\) cannot be \(0\) as \(AB\) is a two-digit number.

Therefore, \(A\) must be \(5\) only. The multiplication is as follows.

\[\begin{align}{5 \;{0}} \\ { \times \;\;{3}} \\ \hline 1\,5\, {0} \\ \hline\end{align}\]

Hence, the values of \(A\),\(B\), and \(C\) are \(5, \,0, \) and \(1\) respectively.