2011 UNCO Math Contest II Answer Key

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1) $41$

2) $\{22,28,30\}$

3) $65$

4) (a) $101$ (b)$\{101, 325, 2501\}$

5) $160$

6) $9$

7) $2^{502}-1504$

8) (a) $74$ (b) $45 \times 74$

9) (a) $T(n+1)+T(n)=\binom{n}{3}$ (b) $T(N) = \binom{N-1}{3}-\binom{N-2}{3}+\binom{N-3}{3}-\binom{N-4}{3}+\cdots$

10) First try $\{1, 2, 3, \ldots , n\}$ for $n= 2, 3, 4, 5$. The crossing off process yields $\{5,23,119,719\}$ each one being one less than a factorial. So for general $n$ you should end up with$(n+1)!-1$. Now look at $n=3$ again and replace $1, 2, 3$ with $a,b,c$ (order does not matter). Crossing off gives you \[(a+b+ab) + c + (a+b+ab)c  =a+b+c+ab+ac+bc+abc\] reminding one of the coefficients in \[(x-a)(x-b)(x-c)= x^3-(a+b+c)x^2+(ab+ac+bc)x-abc\] Now let $x=-1$, and watch what happens remember that $\{a,b,c\} = \{1,2,3\}$. There are other approaches.

11) See solution to #2. Integers that are one less than a prime cannot be written in the form $m +n +m$.

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