Question

Here, $15287$ is divisible by
A.$3$
B.$7$
C.$9$
D.None of these

Answer 1

Hint: The given question can very easily be solved if we know the concept of divisibility tests. A test of divisibility is just a procedure of checking whether the given number is exactly divisible by another number or not without actually working out the process of division. In order to solve this question, we should know the divisibility test rules for number $3,7$ and $9$.

Complete step by step solution:
- Divisibility test for $3$ and $9$: A number is said to be divisible by $3$ if the sum of its digits is divisible by $3$. Similar rule is applicable for $9$ as well. A number is said to be divisible by $9$ if the sum of its digits is divisible by $9$.
- Divisibility test for $7$: If we double the last digit and then subtract it from the remaining leading truncated number, then if the resulting number is divisible by $7$ then it can be said that the original number is also divisible by $7$.
According to the question,
The given number is $15287$.
Let us first test the divisibility for $9$.
We already know that if the sum of digits is divisible by $9$, then the number is also divisible by $9$.
Sum of digits: $1 + 5 + 2 + 8 + 7 = 23$
Clearly,$23$ is not divisible by $9$. Therefore,$15287$ is also not divisible by $9$.
Moreover, as $3$ is a factor of $9$, so we can conclude that even $15287$ is not divisible by $3$.
Now, let us test the divisibility for $7$.
As we already discussed the method of testing the divisibility for $7$, using the same, we get,
$
   \Rightarrow 15287 \to 1528 - 2\left( 7 \right) = 1514 \\
   \Rightarrow 1514 \to 151 - 2\left( 4 \right) = 143 \\
   \Rightarrow 143 \to 14 - 2\left( 3 \right) = 8 \\
 $
Clearly, $8$ is not divisible by $7$ so, $15287$ is also not divisible by $7$.
Therefore, we can conclude that $15287$ is not divisible by either $3$ or $7$ or $9$.

Hence, option (D) is correct.

Note:
Here in this question, we took help of the options to eliminate various other divisibility tests. There is another test for divisibility for $7$i.e., by taking the digits of the number backwards i.e., from right to left and then multiply them progressively by the digits $1,3,2,6,4,5$ and continue repeating this pattern of multipliers until it's necessary. If the sum obtained is divisible by $7$, then the number is divisible by $7$.