Let $$x$$ and $$y$$ be distinct integers where $$1 \leq x \leq 25$$ and $$1 \leq y \leq 25$$. Then, the number of ways of choosing $$x$$ and $$y$$, such that $$x + y$$ is divisible by 5, is _____.

We need to count the number of ways to choose distinct integers $$x$$ and $$y$$ with $$1 \leq x \leq 25$$ and $$1 \leq y \leq 25$$ such that $$x + y$$ is divisible by 5.

In $$\{1, 2, \ldots, 25\}$$, each residue class has exactly 5 elements:

• Residue 0: {5, 10, 15, 20, 25}
• Residue 1: {1, 6, 11, 16, 21}
• Residue 2: {2, 7, 12, 17, 22}
• Residue 3: {3, 8, 13, 18, 23}
• Residue 4: {4, 9, 14, 19, 24}

The valid residue pairs are: $$(0,0), (1,4), (2,3), (3,2), (4,1)$$.

• $$(0,0)$$: Choose $$x$$ and $$y$$ from the same 5 elements, $$x \neq y$$: $$5 \times 4 = 20$$
• $$(1,4)$$: $$5 \times 5 = 25$$
• $$(2,3)$$: $$5 \times 5 = 25$$
• $$(3,2)$$: $$5 \times 5 = 25$$
• $$(4,1)$$: $$5 \times 5 = 25$$

Total ordered pairs:

$$20 + 25 + 25 + 25 + 25 = 120$$

The answer is $$120$$.