Question

Find the digits A and B. [3 MARKS]

$\begin{array}{c}BA\\ \times B3\\ 57A\end{array}$

Answer 1

Concept: 1 Mark
Application: 2 Marks

A can be 0 or 5.

Now the only possible values of B are 1 and 2 $[30\times 30=900]$.

Hence if B is 3, the product of the multiplication will surely be greater than 900 and hence not equal to 575.

When A = 0, either B = 1 or 2, but doesn't satisfy the condition.

When A = 5, B = 2 the product obtained is $25\times 23=575$.

Hence, A = 5 and B = 2.