Question

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 ____

Answer 1

Correct Answer: 120

The correct answer is 120
x+y=5λ
Cases : 

x	y	Number of ways
5λ	5λ	20
5λ+1	5λ+4	25
5λ+2	5λ+3	25
5λ+3	5λ+2	25
5λ+4	5λ+1	25

\(\therefore\) Total number of ways are 120

Answer 2

1. Classify integers modulo 5: The integers from \( 1 \) to \( 25 \) can be grouped based on their residue modulo 5: \[ \{1, 2, 3, 4, 0\} \mod 5. \] Each residue class appears exactly \( 5 \) times (e.g., \( 1, 6, 11, 16, 21 \)). 2. Form pairs where \( x + y \equiv 0 \mod 5 \): For \( x + y \) to be divisible by 5, the residues of \( x \) and \( y \) must satisfy: \[ (x \mod 5) + (y \mod 5) \equiv 0 \mod 5. \] Valid pairs of residues are: \[ (0, 0), (1, 4), (2, 3), (3, 2), (4, 1). \] 3. Count the number of pairs for each case: - Each residue appears \( 5 \) times, so the number of ways to choose a pair for \( (0, 0) \) is: \[ \binom{5}{2} = 10. \] - For other cases, there are \( 5 \cdot 5 = 25 \) pairs for each combination. Total number of pairs: \[ 10 + 25 + 25 + 25 + 25 = 120. \]