Answer only. What is the greatest possible sum of the digits of a six-digit number that is a multiple of 2, 3, 4, 5, 6, 7, 8, 9, 10 and 11?

36

The LCM of 2, 3, 4, 5, 6, 7, 8, 9, 10 and 11 is 2 × 3 × 2 × 5 × 7 × 2 × 3 × 11 = 27720.

A multiple of 27720 will end in 0, and the sum of its digits will be a multiple of 9.

The largest 6-digit number which is multiple of 9 with last digit of 0 is 999990, and the sum of its digits is 45. 45 is not a multiple of 27720, and there is no other way to get a digit sum of 45.

We now look at the next biggest multiple of 9, 9 × 4

= 36.

The largest 6-digit multiple of 27720 is 27720 × 36 = 997920, and the sum of its digits is 36.

answer: a

step-by-step explanation:

10.2 units