Mathcounts National Sprint Round Problems And Solutions -

(\boxed\frac18011)

Thus min sum = 108.

Coordinate geometry turns messy geometry into manageable algebra. Use it liberally. Category 4: Combinatorics – Counting Without Tears Problem (Modeled after 2015 National Sprint #29): How many 4-digit numbers have at least one digit repeated? Mathcounts National Sprint Round Problems And Solutions

(\fraca+bab = \frac317 \Rightarrow 17(a+b) = 3ab). Solve for one variable: (17a + 17b = 3ab \Rightarrow 17a = 3ab - 17b = b(3a - 17) \Rightarrow b = \frac17a3a-17). (\boxed\frac18011) Thus min sum = 108

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. Category 4: Combinatorics – Counting Without Tears Problem

Hidden nuance: A prime number can be the product of 1 and itself, but here ((n+2)(n+7)) is symmetric. If one factor is prime and the other is 1, we already tried. What if one factor is -1 and the other is negative prime? That would give a positive product. Example: (n+2 = -1) → (n=-3) (no). So indeed, no positive (n) works. But the problem exists, so I must have recalled incorrectly. Let’s adjust: A known real problem asks: “Find sum of all integers n such that (n^2+9n+14) is prime.” Answer often is 0 because none exist. But competition problems avoid empty sets.