The number of 3-digit numbers, formed using the digits 2, 3, 4, 5 and 7, when the repetition of digits is not allowed, and which are not divisible by 3 , is equal to ________.

Solution

Maths sir · 4 steps

Count all 60 numbers, then take away the ones whose digit sum is a multiple of 3.

1. Count everything. With 5 digits and no repetition, the hundreds, tens and units places can be filled in 5 × 4 × 3 = 60 ways.
2. Divisibility test. A number is divisible by 3 exactly when its digit sum is. So you only need to know which choice of 3 digits you use, not their order.
3. Find bad triples. Of the C(5, 3) = 10 triples, four have a digit sum that is a multiple of 3: {2,3,4} gives 9, {2,3,7} gives 12, {3,4,5} gives 12 and {3,5,7} gives 15. Each triple can be arranged in 3! = 6 ways, so 4 × 6 = 24 numbers are divisible by 3.

   60 - 4 × 3! = 60 - 24 = 36

4. So. The number of 3-digit numbers not divisible by 3 is 36.

The common slip: 24. Stopping at the count of numbers that are divisible by 3 gives 24. The question asks for the rest, so subtract that count from 60.

The idea to keep. For divisibility by 3 or 9, the order of digits does not matter. Pick the digit sets with the right sum, multiply by the arrangements, and use the complement when the question says "not".

Answer from the NTA final answer key. Solution by PrepWiser.