Solution: Counting Numbers Divisible by 3
We need to count the numbers that satisfy these conditions:
- Range: Greater than 5000 ($N > 5000$) and less than 9000 ($N < 9000$).
- Digits allowed: {0, 1, 2, 5, 9}.
- Repetition: Allowed.
- Divisibility: Must be divisible by 3.
Number Structure
Numbers satisfying $5000 < N < 9000$ must be 4-digit numbers. The first digit ($d_1$) must be 5. The number format is $5d_2d_3d_4$, where $d_2, d_3, d_4$ can be any digit from the set {0, 1, 2, 5, 9}.
Divisibility by 3 Rule
A number is divisible by 3 if the sum of its digits is divisible by 3. The sum is $S = 5 + d_2 + d_3 + d_4$. We require $S \equiv 0 \pmod{3}$.
Since $5 \equiv 2 \pmod{3}$, the condition becomes $2 + d_2 + d_3 + d_4 \equiv 0 \pmod{3}$.
This simplifies to finding combinations where $d_2 + d_3 + d_4 \equiv 1 \pmod{3}$.
Counting Valid Combinations
The remainders of the allowed digits when divided by 3 are:
- Remainder 0: {0, 9} (Count $n_0 = 2$)
- Remainder 1: {1} (Count $n_1 = 1$)
- Remainder 2: {2, 5} (Count $n_2 = 2$)
We need to count the ordered triplets $(d_2, d_3, d_4)$ such that the sum of their remainders modulo 3 is 1. Let $r_2, r_3, r_4$ be the remainders.
Case 1: Remainder sum is 1 (mod 3). The set of remainders is {1, 0, 0}.
The number of ways to arrange these remainders and pick corresponding digits is:
$(n_1 \times n_0 \times n_0) + (n_0 \times n_1 \times n_0) + (n_0 \times n_0 \times n_1)$
$= (1 \times 2 \times 2) + (2 \times 1 \times 2) + (2 \times 2 \times 1) = 4 + 4 + 4 = 12 \text{ ways.}$
Case 2: Remainder sum is 4 (which is equivalent to 1 mod 3). The sets of remainders are {2, 1, 1} and {2, 2, 0}.
- For permutations of {2, 1, 1}:
$(n_2 \times n_1 \times n_1) + (n_1 \times n_2 \times n_1) + (n_1 \times n_1 \times n_2)$
$= (2 \times 1 \times 1) + (1 \times 2 \times 1) + (1 \times 1 \times 2) = 2 + 2 + 2 = 6 \text{ ways.}$
- For permutations of {2, 2, 0}:
$(n_2 \times n_2 \times n_0) + (n_2 \times n_0 \times n_2) + (n_0 \times n_2 \times n_2)$
$= (2 \times 2 \times 2) + (2 \times 2 \times 2) + (2 \times 2 \times 2) = 8 + 8 + 8 = 24 \text{ ways.}$
The total number of valid combinations for $(d_2, d_3, d_4)$ is the sum of ways from Case 1 and Case 2: $12 + 6 + 24 = 42$.
Final Answer
There are 42 numbers that meet all the specified criteria.
Similar Questions
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Important Questions from Algebra
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- Let $\alpha = \frac{-1 + i\sqrt{3}}{2}$ and $\beta = \frac{-1 - i\sqrt{3}}{2}$, $i = \sqrt{-1}$. If $(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}$, then $m$ is _________
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