Question

Find the greatest number 5 digit which on being divided by 9, 12, 24, and 45. leaves 6 as a remainder respectively.​

Answer 1

Answer:

The Greatest 5 digit number which divisible by 9,12,24,45 is 99726.

Step-by-step explanation:

We know that Dividend=Divisor+Quotient+Remainder.

In all cases it leaves 6 as remainder.

STEP 1:

Apply this formula for 9,12,24,45.

STEP 2:

let the 5 digit number be N.              

                 N/9=6⇒  N= 9a+6. ⇒N-6=9a

                 N/12=6⇒  N=12b+6. ⇒N-6=12b

                 N/24=6⇒  N=24c+6. ⇒N-6=24c

                 N/45=6⇒  N=45d+6. ⇒N-6=45d

∴(N-6)is the L.C.M of 9,12,24,45.            (L.C.M of 9,12,24,45=360)

   General form⇒(N-6)=360y.

   N=360y+6.

   we know greatest 5 digit number is 99999.

   Now,99999÷360=277.77

   So,(N-6)=360y

         N-6=360*277=99720.

         N=99720+6=99726.

So,99726 is the largest 5 digit number when divisible by 9,12,24,45 leaves the remainder of 6 in each case.

Hope it was simple and useful.

Good luck and don't forget to upvote if you liked how it was solved.