Question

Which of the following number is divisible by each one of 3, 7, 9 and 11?

• A. $639$
• B. $2079$
• C. $3791$
• D. $37911$
• E. None of these

Answer 1

The correct option is B $2079$

We are given four options out of which we need to find a number that is divisible by each of $3,7,9and11$.

We will have to apply the divisibility tests of each of these numbers on the given numbers to check whether they are divisible or not.

Let's do it one by one:

$\left(i\right)\text{For}3:$

We know that, a number is divisible by $3$, if the sum its digits is divisible by $3$.

The sum of digits for the numbers given in the four options are:

$\left(A\right)639\Rightarrow 6+3+9=18$

$\left(B\right)2079\Rightarrow 2+0+7+9=18$

$\left(C\right)3791\Rightarrow 3+7+9+1=20$

$\left(D\right)37911\Rightarrow 3+7+9+1+1=21$

The sum of digits of $639,2079and37911$ are divisible by $3$. Hence, these numbers are divisible by $3$.

$\left(ii\right)\text{For}7:$

The divisibility test of $7$ states that, a number is divisible by $7$ if the difference between $2$ times the last digit and the rest of the digits is either $0$ or a multiple of $7$.

Let's check for the three numbers that are divisible by $3$.

$\left(A\right)639\Rightarrow 63-2\times 9=45$

$\left(B\right)2079\Rightarrow 207-2\times 9=189$

$\Rightarrow 18-2\times 9=0$

$\left(D\right)37911\Rightarrow 3791-2\times 1=3789$

$\Rightarrow 378-2\times 9=360$

$\Rightarrow 36-2\times 0=36$

We can see that only $2079$ yields $0$. Other two numbers yield $45$ and $36$, which are not multiples of $7$.

Hence, only $2079$ is divisible by $7$.

$\left(iii\right)\text{For}9:$

We know that, a number is divisible by $9$, if the sum its digits is divisible by $9$.

Let's check whether $2079$ is divisible by $9$.

$\left(B\right)2079\Rightarrow 2+0+7+9=18$

$18$ is divisible by $9$. Hence, $2079$ is divisible by $9$.

$\left(iv\right)\text{For}11:$

The divisibility test for $11$ states that, a number is divisible by $11$, if the difference between the sum of digits at odd place and the sum of digits at even places is either $0$ or a multiple of $11$.

Let's check whether $2079$ is divisible by $11$.

$\left(B\right)2079\Rightarrow (2+7)-(0+9)=9-9=0$

The difference is $0$. Hence, $2079$ is divisible by $11$

Hence, $2079$ is divisible by each of $3,7,9and11$.