These problems from the 2008 National Sprint Round demonstrate a steep increase in difficulty.
n=218/2×38/2×54/2×72/2n equals 2 raised to the 18 / 2 power cross 3 raised to the 8 / 2 power cross 5 raised to the 4 / 2 power cross 7 raised to the 2 / 2 power
Reviewing past national-level problems reveals the patterns and creative leaps needed to succeed. Problem 1: Number Theory (Modular Arithmetic) What is the remainder when 320263 to the 2026th power is divided by 11? Solution Strategy: Use Fermat's Little Theorem. Step 1: Identify that 11 is prime. Therefore, Step 2: Divide the exponent by 10. Step 3: Rewrite the expression: Step 4: Simplify modulo 11: Step 5: Calculate Step 6: Divide 729 by 11 to get the final remainder: Answer: 3. Problem 2: Geometry (Area Optimization)
[Read Problem] ──► [First 15 Seconds: Classify & Find Shortcut] │ ├─► Shortcut Found ──► Execute & Double-Check Units │ └─► No Shortcut ─────► Apply Brute Force OR Skip (If Q21-30) Mathcounts National Sprint Round Problems And Solutions
13S=19+227+381+…one-third cap S equals one-nineth plus 2 over 27 end-fraction plus 3 over 81 end-fraction plus …
Pass 2 (Minutes 15–35): Focus on questions 16 through 25. Spend time setting up setups for geometry and combinatorics.
Let’s re-read: “positive integers (n)” and “is a prime number.” If (n=1): (3)(8)=24, not prime. n=2: (4)(9)=36. n=3: (5)(10)=50. n=4: (6)(11)=66. n=5: (7)(12)=84. It seems never prime. These problems from the 2008 National Sprint Round
Mental math and "pencil-and-paper" shortcuts are your only allies.
Provides specialized workbooks that categorize National-level problems by topic. Final Thought
Working Backwards: In many multiple-choice formats, plugging in answers is a viable strategy. However, since MATHCOUNTS is free-response, students must instead use "logical backtracking"—assuming a property is true and seeing if it creates a contradiction. Solution Strategy: Use Fermat's Little Theorem
What is the smallest positive integer n such that n! is divisible by 10¹⁰⁰? Solution: To find the exponent of a prime p in the prime factorization of n!, we use Legendre's Formula. We need 10¹⁰⁰ = 2¹⁰⁰ × 5¹⁰⁰. Since there are fewer factors of 5 than 2 in any factorial, we focus on the exponent of 5.We need the largest power of 5 in n! to be at least 100.Let's approximate: .Let's test n=405: .So, 405 is the smallest integer. Answer: 405 Problem 2: Geometry (Spatial Reasoning)
This is an arithmetic-geometric series. We can solve it elegantly using algebraic manipulation. Let represent the total value of the infinite sum:
— that’s area. But contest answer expected as fraction: ( \frac32 ).