Mathcounts National Sprint Round Problems And Solutions Jun 2026
While the MATHCOUNTS syllabus is broad, the National Sprint Round consistently focuses on four primary pillars of competitive middle school math:
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
For middle school mathematicians, the is the Super Bowl of numbers. At the heart of this prestigious event lies the Sprint Round —a 40-minute, 30-problem gauntlet that tests speed, accuracy, and creative problem-solving. While the MATHCOUNTS syllabus is broad, the National
Sprint Round algebra often involves simplifying complex-looking expressions or solving Diophantine equations (equations where solutions must be integers). Plug into mod 5: 3k+2 ≡ 3 →
Let’s look at three representative problems—one from each difficulty tier.
Triangle BEF: vertices B(8,0), E(3,15), F(24/11, 120/11). Use shoelace formula: Area = 1/2 | x1(y2-y3) + x2(y3-y1) + x3(y1-y2) | = 1/2 | 8(15 - 120/11) + 3(120/11 - 0) + (24/11)(0 - 15) | = 1/2 | 8( (165-120)/11 ) + 3(120/11) + (24/11)(-15) | = 1/2 | 8*(45/11) + 360/11 - 360/11 | = 1/2 | 360/11 | = 180/11.
( 4 + 12 + 36 = 52 ).
