Question:

To find the number of ways of choosing $x$ and $y$, such that $x + y$ is divisible by 5, we proceed as follows: Step 1: Consider the residues mod 5 from 1 to 25. Each residue class (0, 1, 2, 3, 4) has exactly 5 elements. Step 2: We can form pairs $(r, 5-r)$ for $r=1,2$. Each of these pairs gives $5 \times 5 \times 2 = 50$ ordered pairs, since there are 5 choices for $r$ and 5 choices for $5-r$, and the order of $x$ and $y$ matters. Step 3: Next, consider the pairs $(0,0)$. Since $x$ and $y$ must be distinct, there are $5 \times 4 = 20$ ordered pairs of this type. Step 4: Now, consider the pairs $(r,r)$ where $2r \equiv 0 \pmod{5}$. The only possibility is $r=0$, which has already been counted. Step 5: We also need to consider the pairs $(r,r)$ where $r+r \equiv 0 \pmod{5}$. Again, the only possibility is $r=0$, which has already been counted. Step 6: To find the total number of ordered pairs, we add the number of pairs from each of the cases considered: $$ \begin{aligned} \text{Total ordered pairs} &= 5 \times 5 \times 2 \text{ (for $(1,4)$ and $(4,1)$ type)} \\ &\quad + 5 \times 5 \times 2 \text{ (for $(2,3)$ type)} \\ &\quad + 5 \times 4 \\ &= 50 + 50 + 20 \\ &= 120 \end{aligned} $$ Therefore: 120 <div class="key-concept"><strong>Key Concept:</strong> Classify integers 1–25 by residue mod 5; pairs $(x,y)$ with $x+y \equiv 0 \pmod{5}$ come from complementary residue classes</div> <div class="trap-box"><strong>Trap:</strong> Including pairs where $x = y$; the problem states distinct integers</div>