Question

In the following multiplication, each of the different letter denotes a different integer. Each letter stands for the same integer through out. If ${}^{\prime }{B}^{\prime }$ stands for 6 and ${}^{\prime }{E}^{\prime }$ stands for 8, then what is the difference between ${}^{\prime }{F}^{\prime }$ and ${}^{\prime }{D}^{\prime }$ ?
$ABC$
$\times DE$
________________
$ACFB$
$EAGO$
________________
$FHFB$

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

Answer: (B)

$4$

Given that $B=6$ and $E=8$

$A6C$

$\times D8$

________________

$ACF6$

$8AGO$

________________

$FHF6$

________________

As at the Thousand's place $A+8=F$, there is no carry over, thus the value of $A$ can either be $0$ or $1$, but $0$ will make it a 3-digit number. Hence, the value of $A=1$ and thus that of $F=9$. Also, in the first multiplication, $E\times C=B$, i.e., $8\times C=6$, So, ${}^{\prime }{C}^{\prime }$ can either be 2 or 7. By hit and trial, we find that $C=2$. Therefore,

$162$

$\times D8$

________________

$1296$

$+81GO$

________________

$9H96$

________________

As $9+G=9$, therefore, $G$ should be ${}^{\prime }{0}^{\prime }$ which can be obtained when $D=5$, thus the difference between $D$ & $F$ is $F-D=9-5=4$.