When the positive integers a, b, and c are divided by 13, the respective remainders are 9, 7, and 10, respectively. Show that a + b + c is divisible by 13, with calculation.

Let's denote the positive integers a, b, and c as follows:

a = 13k + 9
b = 13m + 7
c = 13n + 10

where k, m, and n are integers.

Now, let's calculate the sum a + b + c:

a + b + c = (13k + 9) + (13m + 7) + (13n + 10)
= 13k + 13m + 13n + 9 + 7 + 10
= 13(k + m + n) + 26
= 13(k + m + n) + 13*2

Since 13(k + m + n) is divisible by 13, we can conclude that a + b + c is also divisible by 13.

Step 1: Convert each number to its modular representation modulo 13:

- a ≡ 9 (mod 13)
- b ≡ 7 (mod 13)
- c ≡ 10 (mod 13)

Step 2: Multiply the modular equations:

```
(a ≡ 9 (mod 13)) × (b ≡ 7 (mod 13)) × (c ≡ 10 (mod 13))
```

Step 3: Calculate the product of the remainders:

```
9 × 7 × 10 = 630
```

Step 4: Find the remainder of the product when divided by 13:

```
630 (mod 13) = 0
```

Step 5: Convert the modular product back to an integer:

```
a × b × c ≡ 0 (mod 13)
```

Step 6: Conclude that a × b × c is divisible by 13 since its remainder when divided by 13 is 0.

Therefore, a × b × c is divisible by 13.