Question

Find the greatest number of five digits which when divided by $ 3,5,8,12 $ will have $ 2 $ as remainder.
 $
  A.{\text{ 99999}} \\
  {\text{B}}{\text{. 99958}} \\
  {\text{C}}{\text{. 99960}} \\
  {\text{D}}{\text{. 99962}} \\
  $

Answer 1

Hint: First find the LCM (least common multiple) of all the given numbers as the divisor where LCM is the least common number or the multiple of two or more numbers. After calculating LCM will take the greatest number and then will divide with the LCM to check whether the given condition has $ 2 $ as the remainder.

Complete step-by-step answer:
Now, find the LCM of the given four numbers by prime factorization method.
 $
\Rightarrow 3 = 3 \times 1 \\
\Rightarrow 5 = 5 \times 1 \\
\Rightarrow 8 = 2 \times 2 \times 2 \\
\Rightarrow 12 = 2 \times 2 \times 3 \\
  $
From the above factorization- (writing the factor once in case of the common multiple)
LCM of all the four numbers is $ = 2 \times 2 \times 2 \times 3 \times 5 $
LCM $ = 120 $
Now, the greatest five digit number is $ 99999 $
Divide the number with the LCM to find the remainder
 $ \Rightarrow \dfrac{{99999}}{{120}} = 39 $
Since, the required remainder is $ 2 $
So, add $ 2 $ and subtract $ 39 $ from the greatest five digit number.
 $ \therefore $ The required number is $ = 99999 - 39 + 2 $
 $ \therefore $ The required number is $ = 99962 $

So, the correct answer is “Option D”.

Note: LCM should be calculated very carefully, as the ultimate answer solely depends on the LCM.LCM can be calculated by the division method also. In the division method write down the given numbers in the row separating them with the comma, divide the number till you reach the prime numbers.