Mathcounts National Sprint Round Problems And Solutions
To understand the rigor of the competition, let us analyze three representative problems inspired by the upper-tier difficulty (Problems 20–30) of historical Mathcounts National Sprint Rounds. Problem 1: Number Theory (Divisibility & Factorization)
Total Divisors=(8+1)(4+1)(2+1)(1+1)=9×5×3×2=270Total Divisors equals open paren 8 plus 1 close paren open paren 4 plus 1 close paren open paren 2 plus 1 close paren open paren 1 plus 1 close paren equals 9 cross 5 cross 3 cross 2 equals 270
a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−(ab+bc+ca))a cubed plus b cubed plus c cubed minus 3 a b c equals open paren a plus b plus c close paren open paren a squared plus b squared plus c squared minus open paren a b plus b c plus c a close paren close paren Mathcounts National Sprint Round Problems And Solutions
Number of divisors=(4+1)⋅(2+1)=5⋅3=15Number of divisors equals open paren 4 plus 1 close paren center dot open paren 2 plus 1 close paren equals 5 center dot 3 equals 15 15 Problem 2: Geometry (Advanced Difficulty) Question: In
(y1+1)+(y2+1)+(y3+1)+(y4+1)+(y5+1)=10open paren y sub 1 plus 1 close paren plus open paren y sub 2 plus 1 close paren plus open paren y sub 3 plus 1 close paren plus open paren y sub 4 plus 1 close paren plus open paren y sub 5 plus 1 close paren equals 10 To understand the rigor of the competition, let
Expect questions on , divisor counts , and GCD/LCM triples . Example: "How many ordered triples 2. Complex Geometry
r=2(2−1)2+1r equals the fraction with numerator 2 open paren the square root of 2 end-root minus 1 close paren and denominator the square root of 2 end-root plus 1 end-fraction The is 30 minutes of pure mathematical intensity
. What is the radius of the smaller circle? Express your answer as a common fraction.
The is 30 minutes of pure mathematical intensity. With 30 problems to solve without a calculator, this round separates the good from the great. It tests not just your math knowledge, but your mental agility, pattern recognition, and ability to perform lightning-fast arithmetic.
∑n=1∞n3n=13+29+327+481+…sum from n equals 1 to infinity of the fraction with numerator n and denominator 3 to the n-th power end-fraction equals one-third plus two-nineths plus 3 over 27 end-fraction plus 4 over 81 end-fraction plus …
The center of the incircle, the center of the smaller circle, and the vertex are collinear, lying on the angle bisector of .The distance from to the center of the large incircle is Set up the Ratio: Let the radius of the smaller circle be . The distance from to the center of the smaller circle is .The distance from to the large center can also be written as the sum of: The distance from to the small center ( The radius of the small circle ( The radius of the large circle ( Therefore:
