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Step-by-Step Solution
Step 1: Interpret the problem
We need to find all distinct ordered pairs (x,y) with 1 \le x \le 25 and 1 \le y \le 25 such that x + y is divisible by 5 (i.e., x + y \equiv 0 \pmod{5} ) and x \neq y .
Step 2: Recognize the possible remainders modulo 5
An integer n when divided by 5 can leave one of the five possible remainders: 0, 1, 2, 3, or 4 . Symbolically, any integer n in the range 1 \le n \le 25 satisfies:
\[
n \equiv r \pmod{5}\quad \text{for some } r \in \{0,1,2,3,4\}.
\]
Step 3: Count how many numbers lie in each remainder class
Between 1 and 25 inclusive, each remainder class modulo 5 has exactly 5 members because 25 \div 5 = 5 . Concretely:
Remainder 0 : \{5, 10, 15, 20, 25\}
Remainder 1 : \{1, 6, 11, 16, 21\}
Remainder 2 : \{2, 7, 12, 17, 22\}
Remainder 3 : \{3, 8, 13, 18, 23\}
Remainder 4 : \{4, 9, 14, 19, 24\}
Step 4: Identify valid remainder pairs
For x + y to be divisible by 5, the pair of remainders (r_x, r_y) must satisfy:
\[
r_x + r_y \equiv 0 \pmod{5}.
\]
The valid combinations of remainders are therefore:
\[
(0,0), \quad (1,4), \quad (2,3), \quad (3,2), \quad (4,1).
\]
Step 5: Count the number of pairs in each valid combination
Case (0, 0):
- Remainder 0 set for x = \{5, 10, 15, 20, 25\} (5 values).
- Remainder 0 set for y = \{5, 10, 15, 20, 25\} (5 values).
- Since x \neq y , each choice of x (5 choices) allows 4 choices for y (all the remaining multiples of 5).
- Hence, number of ordered pairs (x,y) = 5 \times 4 = 20 .
Case (1,4):
- Remainder 1 set for x = \{1, 6, 11, 16, 21\} (5 values).
- Remainder 4 set for y = \{4, 9, 14, 19, 24\} (5 values).
- Since these two sets are disjoint, x \neq y is automatically satisfied for each choice in the sets.
- Hence, number of ordered pairs (x,y) = 5 \times 5 = 25 .
Case (4,1):
- Remainder 4 set for x = \{4, 9, 14, 19, 24\} (5 values).
- Remainder 1 set for y = \{1, 6, 11, 16, 21\} (5 values).
- By similar reasoning, number of ordered pairs (x,y) = 5 \times 5 = 25 .
Case (2,3):
- Remainder 2 set for x = \{2, 7, 12, 17, 22\} (5 values).
- Remainder 3 set for y = \{3, 8, 13, 18, 23\} (5 values).
- Disjoint sets again, so x \neq y automatically for choices from these sets.
- Hence, number of ordered pairs (x,y) = 5 \times 5 = 25 .
Case (3,2):
- Remainder 3 set for x = \{3, 8, 13, 18, 23\} (5 values).
- Remainder 2 set for y = \{2, 7, 12, 17, 22\} (5 values).
- Disjoint sets, so x \neq y is assured.
- Hence, number of ordered pairs (x,y) = 5 \times 5 = 25 .
Step 6: Sum up all valid pairs
The total number of valid ordered pairs (x,y) with x + y divisible by 5 and x \neq y is:
\[
20 + 25 + 25 + 25 + 25 = 120.
\]
Final Answer: 120
