find the greatest number of six digits which on division by 42, 45, 48, 56, and 60 leaves 12 as remainder in each case.

plz answer it fast........

Question

find the greatest number of six digits which on division by 42,
45, 48, 56, and 60 leaves 12 as remainder in each case
plz answer it fast

Duke Biswass · Accepted Answer

Let us write the prime factorisation of the given numbers

42 = 2×3×7

45 = $3^{2}$ ×5

48 = $2^{4}$ ×3

56 = $2^{3}$ ×7

60 = $2^{2}$ ×3×5

LCM of these numbers = product of the highest power of each prime factor

= $2^{4}$ × $3^{2}$ ×5×7

=5040

Logically,any number which is divisible by all 5 above mentioned numbers must be a multiple of their LCM which is 5040

We want the largest 6 digit number of that sort

the largest 6 digit number would be 999999 which when divided by 5040 gives

999999/5040=198.4125

So the largest six digit number divisible by 5040 without remainder is 5040×198 = 997920

when we add remainder 12 to this we get 997932

Hence 997932 is the largest six digit number which when divided by 42,45,48,56,60 gives remainder 12.