Question

A number is divisible by which number, if the number formed by the last three digits of the original number is divisible by $8$.
A. Three
B. Four
C. Eight
D. Seven

Answer 1

Hint: The divisibility rule of number $8$ states that a number is divisible by $8$ if the number formed by the last three digits of the number is divisible by $8$. For example, for the number $15840$, it is quite tough to determine at a glance whether it is divisible by $8$ or not. But using the divisibility rule of $8$, we can easily observe that the number formed by the last three digits of$15840$, i.e$840$ is divisible by $8$ (as $8 \times 105 = 840$), so, according to the statement of divisibility test of$8$, we can say that $15840$ is also divisible by $8$.

Complete step by step answer:
It is given that a number has its last three digits in the form such that the number formed by the last three digits is divisible by $8$. That is the digits at hundredth place, tenth place and units place form a number which is divisible by $8$. So, let us consider the number $N$ of the form,
$N = {a_0} + 10{a_1} + 100{a_2} + 1000{a_3} + ..... + {10^n}{a_n} - - - - \left( 1 \right)$,
where ${a_0}$ is the digit at the unit's place, ${a_1}$ is the digit at tens place, ${a_2}$ is the digit at hundreds place, and so on.

Now, it is given that the last three digits form a number which is divisible by $8$.
Therefore, ${a_0} + 10{a_1} + 100{a_2} = 8k - - - \left( 2 \right)$
Where, k is any constant.
Now, from equations $\left( 1 \right)$ and $\left( 2 \right)$,
$ \Rightarrow N = 8k + 1000{a_3} + {10^4}{a_4} + ..... + {10^n}{a_n}$
Taking $1000$ common from the coefficients of ${a_3},{a_4},{a_5},......,{a_n}$,
$ \Rightarrow N = 8k + 1000[{a_3} + 10{a_4} + 100{a_5} + ..... + {10^{(n - 3)}}{a_n}]$
Taking $8$common from all the terms, we get,
$ \Rightarrow N = 8[k + 125\{ {a_3} + 10{a_4} + 100{a_5} + ..... + {10^{(n - 3)}}{a_n}\} ]$
$ \Rightarrow N = 8C$, where $C = [k + 125\{ {a_3} + 10{a_4} + 100{a_5} + ..... + {10^{(n - 3)}}{a_n}\} ]$
Dividing both sides of the equation by $8$, we get,
$ \therefore \dfrac{N}{8} = C$
Therefore, dividing $N$ by $8$ gives us a constant. Hence, it is divisible by $8$.

Therefore, option C is correct.

Note:The divisibility rules of various numbers help a lot in calculations in any kind of problem, let it be arithmetic, algebra or geometry. The divisibility rules of the numbers are quite simple and easy to remember and pretty quick and fast to implement, large numbers can be identified whether it is divisible by a certain number or not instead of trying by hidden trial method, which saves a lot of time, which is a crucial factor in exams.