How many positive integers less than 1000 have the property that the sum of the digits of each such number is divisible by 7 and the number itself is divisible by 3?

Answer

To solve the problem of finding how many positive integers less than 1000 have the property that the sum of the digits of each such number is divisible by 7 and the number itself is divisible by 3, we can follow these steps:

### Step 1: Understanding the Conditions
We need to find numbers less than 1000 that satisfy two conditions:
1. The sum of the digits (A + B + C) is divisible by 7.
2. The number itself is divisible by 3.

### Step 2: Expressing the Conditions Mathematically
Let’s denote the digits of the number as A, B, and C, where A is the hundreds digit, B is the tens digit, and C is the units digit. The number can be expressed as \(100A + 10B + C\).

From the properties of divisibility:
- A number is divisible by 3 if the sum of its digits is divisible by 3. Therefore, \(A + B + C\) must also be divisible by 3.
- We want \(A + B + C\) to be divisible by both 7 and 3.

### Step 3: Finding Common Multiples
The least common multiple of 7 and 3 is 21. Thus, we can express the sum of the digits as:
\[ A + B + C = 21k \]
for some integer \(k\).

### Step 4: Finding Possible Values for \(A + B + C\)
Since we are looking for positive integers less than 1000, the maximum value for \(A + B + C\) occurs when \(A = 9\), \(B = 9\), and \(C = 9\):
\[ A + B + C \leq 9 + 9 + 9 = 27. \]

Thus, the possible values for \(A + B + C\) that are less than or equal to 27 and divisible by 21 are:
- \(21\) (when \(k = 1\))
- \(42\) (when \(k = 2\), but this exceeds 27)

So, the only feasible value is \(A + B + C = 21\).

### Step 5: Finding Combinations of Digits
Now we need to find the non-negative integer solutions to the equation:
\[ A + B + C = 21 \]
where \(0 \leq A, B, C \leq 9\).

To find the number of solutions, we can use the "stars and bars" combinatorial method. However, we need to account for the restrictions on the digits.

### Step 6: Transforming the Problem
To handle the upper limit of 9 for each digit, we can transform the variables:
Let:
- \(A' = 9 - A\)
- \(B' = 9 - B\)
- \(C' = 9 - C\)

This leads to:
\[ (9 - A') + (9 - B') + (9 - C') = 21 \]
or:
\[ A' + B' + C' = 6 \]
where \(A', B', C' \geq 0\).

### Step 7: Counting the Solutions
Now we need to find the number of non-negative integer solutions to:
\[ A' + B' + C' = 6. \]

Using the "stars and bars" theorem, the number of solutions is given by:
\[ \binom{n + r - 1}{r - 1} = \binom{6 + 3 - 1}{3 - 1} = \binom{8}{2}. \]

Calculating this:
\[ \binom{8}{2} = \frac{8 \times 7}{2 \times 1} = 28. \]

### Conclusion
Thus, the total number of positive integers less than 1000 that satisfy both conditions is **28**.

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