Mathcounts National Sprint Round Problems And Solutions (90% Best)

Use complement counting when “at least one” is cumbersome. Category 5: Advanced Sprint – The Final Four Problems The last 4 problems in a National Sprint Round are notorious. They often combine multiple concepts. Here’s a composite example:

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. Mathcounts National Sprint Round Problems And Solutions

We use identities: ((x+y)^2 = x^2 + 2xy + y^2 \Rightarrow 64 = 34 + 2xy \Rightarrow 2xy = 30 \Rightarrow xy = 15). Use complement counting when “at least one” is

Since (b>0), (3a-17 >0 \Rightarrow a \ge 6). Also integer: (3a-17) divides (17a). Use division: (17a = 17/3*(3a-17) + 289/3) – messy. Instead, rewrite: (b = \frac17a3a-17 = 5 + \frac853a-17) after polynomial division? Let’s check: Divide 17a by (3a-17): quotient 5 (since 5*(3a-17)=15a-85, remainder 2a+85? No, do carefully: (17a) / (3a-17) = 5 + (2a+85)/(3a-17)? That doesn’t help. Better: Set (k = 3a-17), then (a = (k+17)/3), substitute into b: (b = \frac17(k+17)/3k = \frac17k+2893k = \frac173 + \frac2893k). For b integer, (3k) must divide 289 = 17^2. Thus (3k) is a divisor of 289: 1, 17, 289. But 3k positive, k = (3a-17) >0. 3k=1→k=1/3 no. 3k=17→k=17/3 no. 3k=289→k=289/3 no. So no integer k? That means I made an algebraic slip. n=4: (6)(11)=66