Recurring Digits

Logic Level 2

$\begin{array}{ccccc} & & & & 2 &\boxed{\phantom0} \\ \times & & & & \boxed{\phantom0} &3 \\ \hline & & & & \boxed{\phantom0} &3 \\ & & & 2 & \boxed{\phantom0} & \\ \hline & & & 2 & \boxed{\phantom0} &3 \\ \hline \end{array}$

Above shows an incomplete long multiplication. Complete this long multiplication. What is the product of all the missing digits?

The answer is 42.

1 solution

A Former Brilliant Member
Aug 9, 2016

$\begin{array}{ccccc} & & & & 2 &\boxed{1} \\ \times & & & & \boxed{1} &3 \\ \hline & & & & \boxed{6} &3 \\ & & & 2 & \boxed{1} & \\ \hline & & & 2 & \boxed{7} &3 \\ \hline \end{array}$

Therefore, Product of all missing numbers is $6×7=\boxed{42}$ .

Thank youuuuuuuuuuuuu

Pi Han Goh - 4 years, 10 months ago

Recurring Digits

The answer is 42.

1 solution

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