Explain the following proof to me in a much simpler way:
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### Problem Analysis

We need to prove that any integer $$ n > 17 $$ can be expressed as a non-negative integer combination of 3 and 10. That is, for any integer $$ n > 17 $$, there exist non-negative integers $$ x $$ and $$ y $$ such that:

$$ 3x + 10y = n $$

### Approach

This is a classic problem in number theory related to the **Coin Problem**, which deals with the largest monetary amount that cannot be obtained using any combination of coins of specified denominations. Here, the denominations are 3 and 10, which are **coprime** (their greatest common divisor is 1). 

However, 3 and 10 are not coprime (gcd(3, 10) = 1), so the **Frobenius number** (the largest number that cannot be expressed as a combination of these denominations) exists and is given by $$ g(a, b) = ab - a - b = 3 \cdot 10 - 3 - 10 = 17 $$. This means that every number $$ n > 17 $$ can be expressed as a combination of 3 and 10. 

But we will prove this directly without relying on the Frobenius number.

### Direct Proof

We will use the **Chicken McNugget Theorem**, which states that for two relatively prime positive integers $$ a $$ and $$ b $$, the largest integer that cannot be expressed as a combination of $$ a $$ and $$ b $$ is $$ ab - a - b $$. Here, $$ a = 3 $$ and $$ b = 10 $$, and since $$ \gcd(3, 10) = 1 $$, the largest number that cannot be expressed is $$ 3 \cdot 10 - 3 - 10 = 17 $$. 

But we will proceed with a direct proof by induction or by cases.

### Proof by Cases

We can express any $$ n > 17 $$ as follows. Note that:

1. For $$ n = 18 $$: $$ 6 \cdot 3 + 0 \cdot 10 = 18 $$.
2. For $$ n = 19 $$: $$ 3 \cdot 3 + 1 \cdot 10 = 19 $$.
3. For $$ n = 20 $$: $$ 0 \cdot 3 + 2 \cdot 10 = 20 $$.

Now, for any $$ n \geq 18 $$, we can write $$ n $$ as $$ n = 18 + k $$, where $$ k \geq 0 $$. We can then express $$ n $$ in terms of the expressions for 18, 19, and 20 by adding multiples of 3:

- If $$ n \equiv 0 \mod 3 $$, then $$ n = 3x $$, where $$ x \geq 6 $$ (since $$ n \geq 18 $$).
- If $$ n \equiv 1 \mod 3 $$, then $$ n = 19 + 3m = (3 \cdot 3 + 1 \cdot 10) + 3m = 3(3 + m) + 10 \cdot 1 $$, for some $$ m \geq 0 $$.
- If $$ n \equiv 2 \mod 3 $$, then $$ n = 20 + 3m = (0 \cdot 3 + 2 \cdot 10) + 3m = 3m + 10 \cdot 2 $$, for some $$ m \geq 0 $$.

But this is not very systematic. A better approach is to note that we can cover all residues modulo 3:

1. **Case $$ n \equiv 0 \mod 3 $$**:
   - Then $$ n = 3k $$ for some $$ k \geq 6 $$ (since $$ n \geq 18 $$).
   - We can set $$ y = 0 $$ and $$ x = k $$, so $$ 3x + 10y = 3k = n $$.

2. **Case $$ n \equiv 1 \mod 3 $$**:
   - Then $$ n = 3k + 1 $$ for some $$ k \geq 6 $$ (since $$ n \geq 19 $$, and $$ 19 = 3 \cdot 6 + 1 $$).
   - We need to find $$ x, y $$ such that $$ 3x + 10y = 3k + 1 $$.
   - Rewrite as $$ 3x = 3k + 1 - 10y $$, so $$ 1 - 10y \equiv 0 \mod 3 $$, i.e., $$ 10y \equiv 1 \mod 3 $$, i.e., $$ y \equiv 1 \mod 3 $$ (since $$ 10 \equiv 1 \mod 3 $$).
   - Take $$ y = 1 $$, then $$ 3x = 3k + 1 - 10 = 3k - 9 $$, so $$ x = k - 3 \geq 3 $$ (since $$ k \geq 6 $$).
   - Thus, $$ n = 3(k - 3) + 10 \cdot 1 $$.

3. **Case $$ n \equiv 2 \mod 3 $$**:
   - Then $$ n = 3k + 2 $$ for some $$ k \geq 6 $$ (since $$ n \geq 20 $$, and $$ 20 = 3 \cdot 6 + 2 $$).
   - We need to find $$ x, y $$ such that $$ 3x + 10y = 3k + 2 $$.
   - Rewrite as $$ 3x = 3k + 2 - 10y $$, so $$ 2 - 10y \equiv 0 \mod 3 $$, i.e., $$ 10y \equiv 2 \mod 3 $$, i.e., $$ y \equiv 2 \mod 3 $$ (since $$ 10 \equiv 1 \mod 3 $$).
   - Take $$ y = 2 $$, then $$ 3x = 3k + 2 - 20 = 3k - 18 $$, so $$ x = k - 6 \geq 0 $$ (since $$ k \geq 6 $$).
   - Thus, $$ n = 3(k - 6) + 10 \cdot 2 $$.

But we can also handle smaller cases explicitly:

- For $$ n = 18 $$: $$ 6 \cdot 3 + 0 \cdot 10 = 18 $$.
- For $$ n = 19 $$: $$ 3 \cdot 3 + 1 \cdot 10 = 19 $$.
- For $$ n = 20 $$: $$ 0 \cdot 3 + 2 \cdot 10 = 20 $$.
- For $$ n = 21 $$: $$ 7 \cdot 3 + 0 \cdot 10 = 21 $$.
- For $$ n = 22 $$: $$ 4 \cdot 3 + 1 \cdot 10 = 22 $$.
- For $$ n = 23 $$: $$ 1 \cdot 3 + 2 \cdot 10 = 23 $$.
- For $$ n = 24 $$: $$ 8 \cdot 3 + 0 \cdot 10 = 24 $$.
- For $$ n = 25 $$: $$ 5 \cdot 3 + 1 \cdot 10 = 25 $$.
- For $$ n = 26 $$: $$ 2 \cdot 3 + 2 \cdot 10 = 26 $$.
- For $$ n = 27 $$: $$ 9 \cdot 3 + 0 \cdot 10 = 27 $$.
- For $$ n = 28 $$: $$ 6 \cdot 3 + 1 \cdot 10 = 28 $$.
- For $$ n = 29 $$: $$ 3 \cdot 3 + 2 \cdot 10 = 29 $$.
- For $$ n = 30 $$: $$ 10 \cdot 3 + 0 \cdot 10 = 30 $$.
- And so on...

We observe that for $$ n \geq 18 $$, we can always find a combination of 3 and 10 that sums to $$ n $$. The pattern is that we can increment by 3 and cycle through the residues modulo 3, adjusting the number of 10s as needed.

### General Proof

Alternatively, we can use the fact that 10 ≡ 1 mod 3, so we can adjust the number of 10s to get the right residue modulo 3:

1. For any $$ n \geq 18 $$, find $$ y $$ such that $$ n - 10y \equiv 0 \mod 3 $$, i.e., $$ n \equiv 10y \mod 3 $$, i.e., $$ n \equiv y \mod 3 $$ (since $$ 10 \equiv 1 \mod 3 $$).
   - So choose $$ y \equiv n \mod 3 $$, with $$ y \in \{0, 1, 2\} $$.
   - Then $$ n - 10y \equiv 0 \mod 3 $$, and $$ n - 10y \geq n - 10 \cdot 2 = n - 20 $$.
   - But $$ n \geq 18 $$, so $$ n - 20 \geq -2 $$. We need $$ n - 10y \geq 0 $$:
     - If $$ y = 0 $$, $$ n - 10y = n \geq 18 \geq 0 $$.
     - If $$ y = 1 $$, $$ n - 10y = n - 10 \geq 8 \geq 0 $$ (since $$ n \geq 19 $$ when $$ n \equiv 1 \mod 3 $$, because $$ n \geq 18 $$ and $$ n \equiv 1 \mod 3 $$ implies $$ n \geq 19 $$).
     - If $$ y = 2 $$, $$ n - 10y = n - 20 \geq -2 $$. But if $$ n \equiv 2 \mod 3 $$, and $$ n \geq 18 $$, the smallest such $$ n $$ is $$ n = 20 $$, for which $$ n - 20 = 0 \geq 0 $$. For $$ n = 17 + 3 = 20 $$, it works, but $$ n > 17 $$, so $$ n \geq 18 $$. However, $$ n \equiv 2 \mod 3 $$ and $$ n \geq 18 $$ implies $$ n \geq 20 $$ (since $$ 18 \equiv 0 $$, $$ 19 \equiv 1 $$, $$ 20 \equiv 2 $$). So $$ n - 20 \geq 0 $$.

   - Therefore, $$ n - 10y \geq 0 $$ and divisible by 3, so $$ x = \frac{n - 10y}{3} \geq 0 $$.

### Conclusion

We have shown that for any integer $$ n > 17 $$, there exist non-negative integers $$ x $$ and $$ y $$ such that $$ 3x + 10y = n $$. The proof is constructive and provides a way to find $$ x $$ and $$ y $$:

1. Compute $$ r = n \mod 3 $$.
2. Take $$ y = r $$.
3. Compute $$ x = \frac{n - 10y}{3} $$, which is a non-negative integer.

This works because:
- If $$ r = 0 $$, $$ y = 0 $$, $$ x = \frac{n}{3} \geq 6 $$.
- If $$ r = 1 $$, $$ y = 1 $$, $$ x = \frac{n - 10}{3} \geq 3 $$ (since $$ n \geq 19 $$).
- If $$ r = 2 $$, $$ y = 2 $$, $$ x = \frac{n - 20}{3} \geq 0 $$ (since $$ n \geq 20 $$).

Thus, the statement is proven.