Question

How many three-digit numbers would you find which when divided by 3, 4, 5, 6, 7, leave the remainder 1, 2, 3, 4, and 5 respectively?
(A). 4
(B). 3
(C). 2
(D). 1

Answer 1

Hint- In order to solve this question first write what is given to us. This will give us a clear picture of what our approach should be. Here in this question we use the concept of LCM thus we get our desired answer.

Complete step-by-step solution -
Well in this question we should know the very basics LCM and its operation on numbers as we know that LCM is the lowest common multiple here we should also notice some of its operations performed but the base concept is LCM and the consideration of numbers which helped us to find and assume numbers and take examples.
The rest of the question is nothing but basic use operators.
Also one should know we have to look for the remainders.
Here,
 $
  \left( {3 - 1} \right) = 2 \\
  \left( {4 - 2} \right) = 2,\left( {5 - 3} \right) = 2,\left( {6 - 4} \right) = 2{\text{ and }}\left( {7 - 5} \right) = 2 \\
    \\
 $
LCM of 3, 4, 5, 6, 7, is 420.
Consider three-digit number 999
When 999 is divided by 420, the remainder is 159 with quotient 2.
Subtracting 159 from 99 gives 840,
Again subtract 2 from 840=838. (here 2 is subtracted from 840 as the difference between the given divisors and the remainder is 2 in each case)
Now, when 838 is divided by 3, 4, 5, 6, 7 we get the remainders as 1, 2, 3, 4, 5 respectively
Hence only 1 such number is possible.
Therefore, option (D) is the correct option.

Note- Whenever we face such types of questions the key concept is that we should write what is given to us like we did in this question. Well these were the very basics of the question and one should keep in mind the first step and it will help in the interpretation of the whole question. By these basics you should be able to solve the question.