Question

what is the greatest 4 digit number which is exactly divisible by 12 18 40 45

Answer 1

We know that the Greatest number of four digits = 9999


To find the greatest number of 4 digit we have to find the L.C.M of 12, 18, 40,45

Prime factorization :
12= 2×2×3= 2²×3¹

18= 2×3×3= 2¹× 3²

40= 2×2×2×5= 2³×5

45= 3×3×5= 3²×5

L.C.M of 12 ,18,40,&45= 2³×3²×5


L.C.M of 12 ,18,40,&45= 2×2×2×3×3×5

L.C.M of 12 ,18,40,&45= 360

Now divide the largest 4 digit number 9999 by 360

9999 ÷ 360

9999= 360× 27 +279

Remainder= 279

Then Subtract the remainder from 9999

9999- 279= 9720

Hence, greatest number of four digits which is divisible by 12, 18, 40 & 45
= 9720


==================================================================

Hope this will help you....

Answer 2

Solution :-

To find the greatest 4 digit number which is exactly divisible by 12, 18, 40 and 45, first we have to compute the LCM of 12, 18, 40 and 45.

            _______________ 
        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

The LCM of 12, 18, 40 and 45 = 2*2*2*3*3*5 = 360

Now, we will divide the greatest 4 digit number by 360.

The greatest 4 digit number is 9999.
            ______
      360) 9999 (27
              720
            _____
              2799
              2520
           ______
                279 - Remainder
           ______

Now, we have to subtract the remainder, which is 279, from 9999.

⇒ 9999 - 279 = 9720

So, 9720 is the required greatest 4 digit number which is exactly divisible by 12, 18, 40 and 45.

Answer.