Question

Find out $(A+B+C+D)$ such that $AB\times CB=DDD$ where $AB$ and $CB$ are two digit numbers and $DDD$ is a three digit number

• A. $21$
• B. $19$
• C. $17$
• D. $18$

Answer 1

Answer: (A)

$21$

It is given that $AB\times CB=DDD$

Since $DDD$ is a number with same digits so it is better to regard it as a multiple of $111$.

Thus, $DDD=111\times k$

Also, we know that $111=37\times 3$

Therefore,

$DDD=37\times 3k\Rightarrow 3k<28$

Since,

$37\equiv 7\mathrm{mod}10\Rightarrow 3k\equiv 7\mathrm{mod}10\Rightarrow k=9\Rightarrow 3k=27$

Therefore, $37\times 27=999$

Now, we find the value of $A+B+C+D$ as follows:

$A+B+C+D=3+7+2+9=21$

Hence, $A+B+C+D=21$