Mathcounts National Sprint Round Problems And Solutions Jun 2026

Multiple choice? No—Sprint is fill-in-the-blank. But estimation helps avoid wild errors.

Hard — Number theory / modular reasoning Problem: Smallest positive integer n such that n ≡ 2 (mod 3), n ≡ 3 (mod 5), n ≡ 4 (mod 7). Key insight: Solve via CRT. Congruences: n = 3k+2. Plug into mod 5: 3k+2 ≡ 3 → 3k ≡ 1 (mod 5) → k ≡ 2 (since 3 2=6≡1). So k=5t+2 → n = 3(5t+2)+2 = 15t+8. Now mod 7: 15t+8 ≡ 4 → 15t ≡ 3 (mod7). Reduce: 15≡1 (mod7) → t≡3 → t=3 gives n=15 3+8=53. Answer: 53 Mathcounts National Sprint Round Problems And Solutions

When reviewing National Sprint Round solutions, you’ll notice several recurring themes. Mastering these is the secret to a top score. 1. Advanced Combinatorics and Probability Multiple choice

Use estimation and mental shortcuts to avoid time-consuming long-hand arithmetic. Pattern Recognition: Hard — Number theory / modular reasoning Problem: