Mathcounts National Sprint Round Problems And Solutions Page

Now 289 = 17^2. Positive integer factor pairs: (1,289), (17,17), (289,1). Case 1: 3a-17=1 → a=6, then 3b-17=289 → b=102 → sum=108. Case 2: 3a-17=17 → a=34/3 no. Case 3: 3a-17=289 → a=102, then b=6 → same sum 108. Also negative factors? a,b positive so 3a-17> -? Actually if a=1, 3-17=-14, product with negative to get 289, but then b negative. So only positive pairs.

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

Then (x^3 + y^3 = (x+y)(x^2 - xy + y^2) = 8 \cdot (34 - 15) = 8 \cdot 19 = 152). Now 289 = 17^2

Let’s instead take a from 2018 National Sprint #22: How many positive integers (n) less than 100 have exactly 5 positive divisors? Case 2: 3a-17=17 → a=34/3 no

The factors could be -1 and -prime? But (n>0) gives positive factors. So no solutions? That can’t be – the problem expects a sum.