AKASH GOYAL AskiitiansExpert-IITD
Last Activity: 13 Years ago
Dear Darshita
If a number is divisible by 3, the sum of its digits is divisible by 3. We need to choose five digits from {0,1,2,3,4,5} which add to a multiple of 3. We could choose:
{1, 2, 3, 4, 5}
or
{0, 1, 2, 4, 5}
If we use the digits {1, 2, 3, 4, 5}, we have 5 choices for the first digit, 4 for the second, etc- 5! = 120 numbers we can make in total.
If we use the digits {0, 1, 2, 4, 5}, we only have 4 choices for the first digit (because it cannot be zero), 4 choices for the second digit, 3 for the third, etc- 4*4! = 96 numbers we can make in total.
120+96 = 216.
All the best.
AKASH GOYAL
AskiitiansExpert-IIT Delhi
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