Question

Find the greatest number of 6 digits which on dividing by 15,18,21,24 and 27 leaves 10, 13, 16, 19 and 22 as remainders.

Answer 1

Solution:-

Lcm of (15,18,21,24,27) = 3*3*2*5*7*4*3=7560

If we take the difference between the given number and remainder of the 6 digit number we get,

(15 - 10) = 5

(18 - 13) = 5

(21 - 16) = 5

(24 - 19) = 5

(27 - 22) = 5

So, difference will be same in every time.

we know that largest 6 digit number is 999999.

if we divided 999999 by the lcm 7560 ,we get a remainder which is 2079.

if we divided 999999 by the lcm 7560 ,we get a remainder which is 2079.

Now, if we subtract remainder 2079 from 999999 we get largest 6 digit number, which is divisible by given number

(999999 - 2079) = 997920[ but required number gives difference in every time]

required greatest 6 digits number is (997920 - 5)[ 5 is the difference ] = 997915

Hence,the greatest number of 6 digits is 997915 which on dividing by 15,18,21,24 and 27 leaves 10, 13, 16, 19 and 22 as remainders.(Ans)

Answer 2

Answer:

Step-by-step explanation: