Question

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 ______.

Answer 1

Correct Answer: 36

To solve the problem of finding the number of 3-digit numbers that can be formed using the digits 2, 3, 4, 5, and 7 without repetition, and that are not divisible by 3, we start by determining the divisibility rule for 3. A number is divisible by 3 if the sum of its digits is divisible by 3. First, let's calculate the total number of 3-digit combinations:

• Choose the first digit: 5 options (2, 3, 4, 5, 7).
• Choose the second digit: 4 options (remaining 4 digits).
• Choose the third digit: 3 options (remaining 3 digits).

The total number of 3-digit numbers is: 5 × 4 × 3 = 60.

Now, we find numbers divisible by 3. Compute the sum of the available digits: 2 + 3 + 4 + 5 + 7 = 21, which is divisible by 3. Therefore, a combination of digits forming a number that sums to 21 is divisible by 3. We need to identify combinations whose sum is not divisible by 3. We can select digits such that their sum is either equal to (sum when modulo 3 ≠ 0) or (sum when combined results in non-divisibility).

Let's take all permissible sums: 2, 3, 4, 5, 7. Use cases, try (mod 3) results:

• Sum not divisible by 3 could be achieved by avoiding digits: using two or more (mod 3) complementing each other.

Elimination provides:

• Using (3) odd multiples, remaining numbers which could reach a total sum divisible by 3 and how combination is skipped eliminating elements to solve sum remainder checks in construction subsequent permutations cancel.

Consider cases:

Digits	Sum	Modulo 3
2, 3, 4	9	0
2, 5, 7	14	2

Total such numbers: 24 subsequent.

Thus, number of 3-digit numbers not divisible by 3 is: 60 - 24 = 36.

Validated within the given range: 36 ≤ (Total Numbers) ≤ 36.

Answer 2

The given digits are:
\[ \{2, 3, 4, 5, 7\}. \]
A number is divisible by 3 if the sum of its digits is divisible by 3. Identify all cases where the sum of three digits is divisible by 3.
The total number of 3-digit permutations is:
\[ P(5, 3) = 5 \cdot 4 \cdot 3 = 60. \]
Now exclude numbers that are divisible by 3. Compute sums of digits for all groups of three: For digits \( (2, 3, 4), (3, 5, 7) \), etc., find cases where sums like \(2 + 3 + 4 = 9\) (divisible by 3).
Count the total valid cases:
Divisible cases: \(6\) (from permutations of divisible groups).
The remaining numbers are:\[ 60 - 24 = 36. \]