Question

In the following multiplication, each of the letters denote a different integer. Each letter stands for the same integer throughout. the question, where A stands for 2 and E stands for 6.

A B A

$\times$ C A

----------------

D A D

E C E X

--------------------

E F G D

What is the value of B${}^2\times$ D${}^3$?

• A. $16$
• B. $64$
• C. $8$
• D. $256$

Answer 1

Answer: (B)

$64$

Given:

A B A

$\times$ C A
D A D
E C E X
E F G D

As $A=2,E=6$.

2 B 2
x C 2
D 2 D
6 C 6 X
6 F G D

In first multiplication,
$2\times 2=$ $D$ $=4$ and

$2\times$ $B$ $=2$ (unit digit)

So $B$ can be either $1$ or $6$,

if it is $B$ $=6$, then the product is $2\times 6=12$, where $1$ is a carry over but, at the Hundred's place also $2\times 2=4$.

Thus $2\times$ $B$ $=2$ and not $12$ where $B=1$.

In second multiplication.
$C$ $\times 2=6$ so $C$ is either $3$ or $8$,

if it $C=8$,

then the Ten's place multiplication would have become $B\times C+1=1\times C+1$.

but it remains $1\times C=C$, thus $C=3$.

Therefore,
2 1 2
$\times$ 3 2
4 2 4
6 3 6 x
6 7 8 4

Now,

$D=4$

$B=1$

$B$ ${}^2\times$ $D$ ${}^3=1^2\times 4^3=64$