Question

In the following multiplication, each of the letters denote a different integer. Each letter stands for the same integer throughout 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

Then, $D$ $E$ $A$ $F$ stands for which of the following?

• A. $6784$
• B. $4627$
• C. $3247$
• D. $3216$

Answer 1

Answer: (B)

$4627$

Given:

A B A

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

As $A=2$ and $E=6$

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,

$D=4$

$E=6$

$A=2$

$F=7$

$D$ $E$ $A$ $F$ $=4627$