Problem (Mock National Level):A bag contains 5 red marbles and 5 blue marbles. If three marbles are drawn at random without replacement, what is the probability that at least two are red?
The first 20 problems are typically easier; solve them quickly to bank time for the harder final 10. Mental Math:
To excel in the National Sprint Round, top competitors employ specific tactical approaches:
Let n be a positive integer less than or equal to 1000. If the last two digits of n are reversed, the resulting integer is exactly 85 percent of n. What is the sum of the possible values of n?
The "First 10" Sprint: Elite competitors aim to finish the first 10 problems in under 5 minutes. These are generally straightforward and serve as a "warm-up" to save time for the grueling final five problems. Mathcounts National Sprint Round Problems And Solutions
Strategy: Always look for factoring patterns before brute force.
22−2=r(2+1)2 the square root of 2 end-root minus 2 equals r open paren the square root of 2 end-root plus 1 close paren
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.
16−8r+r2+9=r216 minus 8 r plus r squared plus 9 equals r squared 25−8r=025 minus 8 r equals 0 r=258r equals 25 over 8 end-fraction The center of the circle is and its radius is 25825 over 8 end-fraction . The standard equation of this circle is: Problem (Mock National Level):A bag contains 5 red
N=35(9x+4)+33cap N equals 35 open paren 9 x plus 4 close paren plus 33 N=315x+140+33cap N equals 315 x plus 140 plus 33 N=315x+173cap N equals 315 x plus 173 The general solution is . The smallest positive integer solution is when , which gives , it satisfies all conditions of the problem. Problem 3: Geometry (Power of a Point & Right Triangles) Problem: In right triangle ABCcap A cap B cap C . A circle is tangent to side ABcap A cap B and passes through the midpoint of hypotenuse ACcap A cap C . If the circle intersects side BCcap B cap C at a second point , find the length of segment BDcap B cap D
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)
(\boxed4464)
This article provides a deep dive into the structure, strategy, and specific problem-solving techniques required for the Sprint Round. We will analyze real problem types from past nationals, walk through detailed solutions, and offer a tactical blueprint to boost your speed and accuracy. Mental Math: To excel in the National Sprint
Note: These are sample-style problems that reflect Sprint characteristics; each solution focuses on the key insight and an efficient path rather than lengthy exposition.
Subtract 5 from both sides: $2x = 6$. Divide both sides by 2: $x = 3$.
⌊20/2⌋+⌊20/4⌋+⌊20/8⌋+⌊20/16⌋=10+5+2+1=18the floor of 20 / 2 end-floor plus the floor of 20 / 4 end-floor plus the floor of 20 / 8 end-floor plus the floor of 20 / 16 end-floor equals 10 plus 5 plus 2 plus 1 equals 18 So, 2¹⁸ divides 20!.