Complementary Counting

Divisible by 2 and 3 means the number should also be divisible by 6 and it should not be divisible by 4 and 9.

Smallest 3-digit multiple of 6 = 102

Largest = 996

Number of multiples: $$\dfrac{\left(996-102\right)}{6}+1=150$$

Now, we need to subtract the numbers divisible by both 4 and 9 (i.e., multiples of 36)

Smallest = 108, Largest = 972

$$\dfrac{\left(972-108\right)}{36}+1=25$$

Final answer: 150 - 25 = 125

We need 5-digit numbers greater than 50000 using digits 0, 1, 2, 3, 4, 5, 6, 7 (repetition allowed), where the sum of the first and last digits is at most 8.

Since the number must be greater than 50000, the first digit $$a$$ must be 5, 6, or 7.

The middle three digits can each be any of the 8 digits (0 through 7), giving $$8^3 = 512$$ choices.

For the last digit $$e$$, we need $$a + e \leq 8$$ where $$e \in \{0, 1, 2, 3, 4, 5, 6, 7\}$$.

Case 1: $$a = 5$$: then $$e \leq 3$$, so $$e \in \{0, 1, 2, 3\}$$ — 4 choices.

Case 2: $$a = 6$$: then $$e \leq 2$$, so $$e \in \{0, 1, 2\}$$ — 3 choices.

Case 3: $$a = 7$$: then $$e \leq 1$$, so $$e \in \{0, 1\}$$ — 2 choices.

Total number of such 5-digit numbers = $$(4 + 3 + 2) \times 512 = 9 \times 512 = 4608$$

However, the question asks for numbers strictly greater than 50000. The number 50000 itself satisfies all conditions (first digit 5, last digit 0, sum = 5 ≤ 8), but 50000 is not greater than 50000. So we must exclude it.

Therefore, the required count = $$4608 - 1 = 4607$$

Total 3-digit numbers (no repetition): $$^5P_3=5\times4\times3=60$$

Now, for a number to be divisible by 3, the sum of its digits must be divisible by 3.

So, classifying digits by mod 3:

Mod 0: {3}

Mod 1: {4, 7}

Mod 2: {2, 5}

For sum ≡ 0 (mod 3), possible valid combination:

(0, 1, 2)

Number of ways of choosing 1 from each group:   $$1\times2\times2=4$$ ways (digit sets)

Each set can be arranged in: $$3!=6$$ ways

So total numbers divisible by 3:$$4\times6=24$$