Question

Find the largest number of four digit which is exactly divisible by 12 15 18 and 27

Answer 1

To find the greatest number divisible by 12, 15, 18 and 27, we will first find their LCM.


_______________ 
        2  | 12, 18, 40, 45
            |_______________
        2  |   6,  9,  20, 45 
            |_______________
        2  |   3,  9,  10, 45
            |_______________
        3  |   3,  9,   5,  45 
            |_______________
        3  |   1,  3,   5,  15
            |_______________
        5  |   1, 1,    5,   5
            |_______________
            |   1, 1,    1,   1


LCM = 2 × 2 × 2 × 3 × 3 × 5

LCM = 360


The greatest four digit = 9999


Now we will divide the greatest number by the LCM, i.e, 9999 ÷ 360

______
      360) 9999 (27
              720
            _____
              2799
              2520
           ______
                279 - Remainder
           ______

We got remainder as 279


So we will subtract 279 from 9999

9999 - 279 = 9720



$\therefore{}$ The greatest number divisible by 12, 15, 18 and 27 is $\bf{9720}$

Answer 2

here is your answer OK ☺☺☺☺☺


it has two method to solve.......

first.........


The Largest number of four digits is 9999.

Required number must be divisible by L.C.M. of 12,15,18,27 = 540.

On dividing 9999 by 540, we get 279 as remainder .

Required number = (9999-279) = 9720.

Hence, required largest number is 9720.

I hope you got your answer…


note : why we lcm method use...

we are asked to find a number which is divisible by a set of numbers( a, b , c, d,... ) then that number is also divisible by their Least Common Multiplier (LCM).

OK I hope I help you