Integers m and n are chosen uniformly and independently from {1, 2, …, 100}. What is the probability that 7^m + 7^n is divisible by 5?

Introduction / Context:
We need to evaluate 7^m + 7^n modulo 5. Working modulo 5 simplifies because powers of 7 reduce to powers of 2 (since 7 ≡ 2 mod 5) with a period of 4.

Given Data / Assumptions:

• m and n are independently uniform on {1, …, 100}.
• 7 ≡ 2 (mod 5).
• Cycle of 2^k (mod 5) has period 4: 2, 4, 3, 1 for k ≡ 1, 2, 3, 0 (mod 4).

Concept / Approach:
Let f(r) be 2^r mod 5 with r in {1,2,3,0}. Then 7^m + 7^n ≡ f(m mod 4) + f(n mod 4) (mod 5). Count residue pairs that sum to 0 mod 5.

Step-by-Step Solution:
Residue values by exponent class: {2, 4, 3, 1}.Pairs summing to 0 mod 5 are (2,3), (3,2), (4,1), (1,4) → 4 favorable pairs.Since m mod 4 and n mod 4 are each uniform over 4 classes, total pairs = 16, favorable = 4.Probability = 4/16 = 1/4.

Verification / Alternative check:
Because 100 is a multiple of 4, residues are exactly balanced (25 each), justifying uniformity on residue classes.

Why Other Options Are Wrong:
1/7 and 1/49 ignore the 4-cycle structure; 1/8 undercounts; 3/8 overcounts.

Common Pitfalls:
Forgetting that exponent classes 0..3 map to specific residues of 2^k (mod 5) and mixing the class labels.

Final Answer:
1/4