Question

The largest 3- digit number divisible by 19 is
A. 988
B. 998
C. 999
D. 969

Answer 1

Hint: The largest 3 digit number is 999. So first find whether 999 is divisible by 19 or not. If it is divisible by 19 then 999 is our required number. If it is not divisible by 19, then note the remainder of the division and subtract that remainder from 999 to get the required number.

Complete step by step solution:
Here the dividend is 988. Dividend must always be equal to the sum of (product of quotient and divisor) and remainder. This theorem is called remainder theorem
We are given to find the largest 3- digit number divisible by 19.
The largest 3- digit number possible is 999 as the next number is 1000 which is a 4- digit number.
So first we are dividing 999 by 19, $ \dfrac{{999}}{{19}} $ . 999 is not divisible by 19 and 11 is the remainder we get after the division.
So to find the largest 3-digit number divisible by 19 subtract 11 from 999.
 $\Rightarrow 999 - 11 = 988 $ is our required number.
When 988 is divided by 19, the quotient is 52 and the remainder is obviously zero.
Hence, the correct option is Option A, 988.
So, the correct answer is “Option A”.

Note: Another approach.
First we need to find the smallest 4-digit number.
The smallest four digit number is 1000.
So first we are dividing 1000 by 19, $ \dfrac{{1000}}{{19}} $ . 1000 is not divisible by 19 and 12 is the remainder we get after the division. So subtract 12 from 1000 to get the largest 3-digit number divisible by 19.
 $ 1000 - 12 = 988 $ is our required number.