CBSE Class 9 Answered

In the episode we have aplied Fundamental theorem of Arithmetic and not Euclid's theorem.

Greatest number of 6 digits = 999999

we need a number which is divisible by all of 24,  15 and 36

Number which is divisible by 24,15 and 36 is also divisible by the L.C.M. of 24, 15, and 36.

The prime factorization of  24 = 2³x3

15=5x3
36 = 3² x 2²

L.C. M. = Product of greatest power of each prime factor involved in numbers=8x9x5 = 360

      now 999999 is not divisible by 360

Hence required number = 999999 – remainder  when 999999 divided by 360

Required No is 999999 –279 = 999720.