Grades
distribution
90-100 A (7)
65-89 B (8)
45-64 C (3)
35-44 D (2)
median= 82
most common score =100
1. (a)
Since $GCD(2,n)=1$, if $n$ were prime we would have $2^{n-1}\equiv 1\mod n$, not 4.
So $n$ cannot be prime.
(b) Reducing modulo 3 we get $n$, $n+1$ and $n+2$, three consecutive integers. So one of them must
be divisible by 3, and they are all at least 4, so that is not prime.
2. (a) For even parts distinct we have 5,41,32,311,2111,11111 (6 total).
For multiplicity at most 3 we have 5,41,32,311,221,2111 (6 total).
(b) For $A(x)$, if a part $k$ has multiplicity at most 3, we have
$$
1+x^k+x^{2k}+x^{3k}=\frac{1-x^{4k}}{1-x^k}
$$
So
$$
A(x)=\prod_{k=1}^\infty (1+x^k+x^{2k}+x^{3k})=\prod_{k=1}^\infty \frac{1-x^{4k}}{1-x^k}.
$$
For $B(x)$, we can use odd parts with no restriction and even parts only dstinct,
$$
B(x)=\prod_{k=1}^\infty \frac{1+x^{2k}}{1-x^{2k-1}}.
$$
Factoring $1-x^{4k}=(1+x^{2k})*(1-x^{2k})$ in $A(x)$ yields $B(x).$
3. (a) the Catalan number $C_7=429.$
(b) From (0,0) to (3,3) we get $2*C_2$ (below subdiagonal or above superdiagonal),
(3,3) to (7,7) we get
$2*C_3$ so $4*C_2*C_3.$
4. The number of all possible blockwalks is the left side $\binom{2n}{n}.$ Lets classify these blockwalks by the last time they hit the diagonal before $(n,n).$ If this point is $(k,k)$, where $0\le k\le n-1$, there are $\binom{2k}{k}$ posible blockwalks from $(0,0)$ to $(k,k)$. The remaining blockwalk from $(k,k)$ to $(n,n)$ may not intersect the diagonal anywhere else. So it must be below the subdiagonal or above the superdiagonal, there are $2*C_{n-k-1}$ such blockwalks.
5. (a) $B_2=5$ ($\varnothing, 1, 11, 2, 21$), and a good Catalan guess is $B_3=14$
(b)
In fact $B_n=C_{n+1}$. The border of a Ferrers diagram of such a partition gives a blockwalk
from $(0,0)$ to $(n+1,n+1)$ which stays at or above the diagonal.