Question

Which of the following numbers is divisible by $99$

• A. $913462$
• B. $114345$
• C. $135792$
• D. $3572406$

Answer 1

The correct option is B

$114345$

Step 1. Explanation for the correct option.

The number is divisible by $99$ if it is divisible by both $9$ and $11$.

• Consider $114345$.

Divisibility by $9$.

Sum of digits

$$\begin{gathered} =1+1+4+3+4+5 \\ =18 \end{gathered}$$

And $18$ is divisible by $9$.

Hence the number $114345$ is divisible by $9$.

Divisibility by $11$.

Difference of the sum of alternate digits

$$\begin{gathered} =(1+4+4)-(1+3+5) \\ =9-9 \\ =0 \end{gathered}$$

Therefore the number $114345$ is divisible by $11$

Since the number is divisible by both $9$ and $11$.

Hence, the number is divisible by $99$.

Step 2. Explanation for the incorrect options.

• Consider $913462$.

Divisibility by $9$.

Sum of digits

$$\begin{gathered} =9+1+3+4+6+2 \\ =25 \end{gathered}$$

We know that $25$ is not divisible by $9$. Hence the number $913462$ is divisible by $9$.

Therefore it is not divisible by $99$.

• Consider $135792$.

Divisibility by $9$.

Sum of digits

$$\begin{gathered} =1+3+5+7+9+2 \\ =27 \end{gathered}$$

And $27$ is divisible by $9$.

Hence the number $135792$ is divisible by $9$.

Divisibility by $11$.

Difference of the sum of alternate digits

Difference of the sum of alternate digits

$$\begin{gathered} =(1+5+9)-(3+7+2) \\ =15-12 \\ =3 \end{gathered}$$

And $3$ is not a multiple of $11$. Therefore the number $135792$ is not divisible by $11$.

Hence, the number is not divisible by $99$.

• Consider $3572406$

Divisibility by $9$.

Sum of digits

$$\begin{gathered} =3+5+7+2+4+0+6 \\ =27 \end{gathered}$$

And $27$ is divisible by $9$. Hence the number $3572406$ is divisible by $9$.

Divisibility by $11$.

Difference of the sum of alternate digits

$$\begin{gathered} =(3+7+4+6)-(5+2+0) \\ =20-7 \\ =13 \end{gathered}$$

And $13$ is not a multiple of $11$. Therefore the number $3572406$ is not divisible by $11$.

Hence, the number is not divisible by $99$.

Hence, the correct option is $\mathrm{B}$.