find the greatest number of 3 digits which is divisible by 18,24, and 36

Question

use the grid method

Answer 1

18=2×3×3; 24=2×2×2×3; 36=2×2×3×3.

So 18 is the product of one 2 and two 3s; 24 is the product of 3 2s and one 3; 36 is the product of two 2s and two 3s. The largest number of twos therefore is 3; the largest number of threes is 2.

So they will all divide into number which is the product of 3 twos and 2 threes.

This number is 2×2×2×3×3=8×9=72. But this has only two digits. We need a multiple of 72 which is the largest containing 3 digits. The biggest 3-digit number is 999, which is not a multiple of 72. The largest multiple of 72 is 13 because 999/72=111/8=13⅞ (14×8=112).

	70	2
10	700	20
3	210	6

So add the boxes in the grid together: 910+26=936. This is the largest 3-digit number into which 18, 24 and 36 all divide.