Question

The greatest number of $4$- digits exactly divisible by $12,15,20\& 35$ is
$
  A.9999 \\
  B.9660 \\
  C.9832 \\
  D.9860 \\
 $

Answer 1

Hint- In order to solve such a question consider the largest number of given number of digits and then find the number to be subtracted from the largest number to make it an integral product of LCM of the numbers.

Complete step-by-step answer:

Given that we need to find the largest 4 digit such number exactly divisible by $12,15,20\& 35$.

As we know that largest 4 digit number is$ = 9999$

LCM can be found out by the prime factorization method.

Finding prime factorization of each of the number $12,15,20\& 35$
$
  12 = 2 \times 2 \times 3 \\
  15 = 3 \times 5 \\
  20 = 2 \times 2 \times 5 \\
  35 = 5 \times 7 \\
 $

LCM of $12,15,20\& 35 = 2 \times 2 \times 3 \times 5 \times 7$

LCM of $12,15,20\& 35 = 420$.

On dividing $9999$ by $420$ , we obtain 339 as remainder.

So in order to make a 4 digit number to be integral multiple, we must subtract this remainder from the dividend.

Therefore largest 4 digit number exactly divisible by $12,15,20\& 35$ will be
$
   = 9999 - 339 \\
   = 9660 \\
 $

Hence $9660$ is the required number and option B is the correct option.

Note- In order to find the common multiple of given numbers LCM is the way to proceed. The method shown above is the easiest. The question can also be done by checking for greatest number after checking for greatest number and continuing. In similar questions as in above when we need to find the smallest number, proceed in the same way but in that case just add a number in order to make the integral multiple.