Non-zero coefficients

Terms in the expansion of $(f(x)$ look like: $(x^{20})^i \cdot (x^{13})^j \cdot 1^k$ where $i+j+k = 33$ .

So we need to find all the possible distinct values of $20i + 13j$ subject to the constraints $0 \le i+j \le 33$ and $0 \le i,j$ .

Now, if we restrict $j$ to the range $[0, 19]$ then we still include all possible values. This is because if there is some $N = 20i + 13j$ with $j \ge 20$ , then the value $N = 20(i+13) + 13(j-20)$ is in the restricted range already, and $i+13+j-20 = i+j-7$ satisfies the constraint $0 \le i+j \le 33$ .

Since $13$ and $20$ are coprime, we do not have any duplicate values of N in this range either.

Now, if $j=0$ then $i \in [0, 33]$ , i.e. 34 possibilities.

And if $j=1$ then $i \in [0,32]$ , i.e. 33 possibilities.

Carry on to $j=19$ implying $i \in [0,14]$ which is 15 possibilities.

So the sum we want is $34 + 33 + .... + 15 = \boxed{490}$

The first thing to note is that all nonzero terms upon expansion (even before combining like terms, if you were to distribute everything out) will have positive coefficients, so we do not have to worry about two like terms simplifying to 0. Essentially, the question is asking for the number of distinct exponents that can be formed upon expansion (we will ignore terms with a coefficient of 0).

Thus, we will consider only the exponents and not the coefficients. Upon expansion (you may wish to imagine using the Multinomial Theorem if you cannot easily understand the following), all exponents will have the form $20a + 13b + 0c$ , where $a + b + c \le 33$ and $a, b, c \in \mathbb{W}$ . We have $0c$ to account for the constant in the factored form, but for our purposes, it is redundant, so we reduce the problem to finding the number of distinct values that can be formed from the expression $20a + 13b$ where $a + b \le 33$ .

To do this, we will overcount and then subtract the overcounting. If we just consider the number of nonnegative integer pairs $(a, b)$ while $a + b \le 33$ , we get $\binom{35}{2} = 595$ . If this is not clear, consider each case (eg: $a + b = 0$ has $1$ solution, up to $a + b = 33$ has $34$ solutions) and add them up.

Now we add the restriction of the expression $20a + 13b$ being distinct for each $(a, b)$ . Since $20$ and $13$ are relatively prime, the first case of overcounting deals with the fact that $20 \cdot 13 = 13 \cdot 20$ , which occurs from the pairs $(13, 0)$ and $(0, 20)$ . In fact, all cases of overcounting will occur from the pairs $(13n+m, k)$ and $(m, 20n+k)$ , where $k, m,n \in \mathbb{N}$ . However, it is clear that for any $n > 1$ , we have $a + b > 33$ , so we only need to consider the case where $n = 1$ . We see that we must have the condition $m + k \le 13$ , and there are $\binom{15}{2} = 105$ ways for this to be satisfied.

Therefore, there will be $595 - 105 = \boxed{490}$ terms with a nonzero coefficient.

We can write $f(x)$ as the product of $33$ brackets: $(x^{20}+x^{13}+1)(x^{20}+x^{13}+1)(x^{20}+x^{13}+1)....(x^{20}+x^{13}+1)$

A term in the expansion of $f(x)$ is obtained by choosing one term from each of the $33$ brackets and multiplying these terms together. Let's say the term $x^{20}$ is chosen $a$ times and the term $x^{13}$ is chosen $b$ times. The power of the term obtained will be $20a+13b$ .

The problem thus reduces to the following. Given two non-negative integers $a$ and $b$ such that $a+b\leq 33$ , how many distinct values can $20a+13b$ take?

We will solve this by first counting the total number of pairs of $a$ and $b$ , then subtracting the number of times two pairs give repeated values of $20a +13b$ .

We start by counting the number of pairs of non-negative integers $a$ and $b$ such that $a+b \leq33$ . When $a=0$ , $b$ can range from $0$ to $33$ , giving $34$ possible pairs. When $a=1$ , $b$ can range from $0$ to $32$ , giving $33$ possible pairs.When $a=2$ , $b$ can range from $0$ to $31$ , giving $32$ possible pairs. Continuing in this pattern, we will reach the case when $a=33$ , $b$ can only be $0$ , giving 1 possible pair. Thus the total number of pairs is $34+33+32+...+1= \binom{35}{2}=595$

Now we count the cases where $20a+13b$ are not distinct. Consider the case when $20a+13b=20a'+13b'$ , with $a' \gt a$ (and therefore $b' \lt b$ . This happens if and only if $a' = a+13t$ and $b' =b-20t$ for some positive integer $t$ . We have $0 \leq b'=b-20t \leq 33-20t$ so $33-20t \geq 0$ , giving $t \leq 1$ . Since $t$ is positive, $t$ can only be $1$ .

Thus we see that $a'=a+13$ and $b'=b-20$ . We then proceed to count the number of duplicates. For convenience, we denote $(a,b) \equiv (a',b')$ to show that this is a pair of duplicates.

When $a=0$ we have the following duplicates: $(0,20) \equiv (13,0), (0,21) \equiv (13,1), (0,22) \equiv (13,2),...(0,33) \equiv (13,13)$ Altogether there are $14$ pairs

Continuing in the same way, we see that when $a=1$ there are $13$ pairs, when $a=2$ there are $12$ pairs,... until when $a=13$ there is $1$ pair.

Thus the total number of duplicates is $14+13+12+...+1 = \binom{15}{2} = 105$

The answer we need is thus equal to $595-105=\fbox{490}$

We can find the total number of terms since we can each of the 33 "powers" into a "20" box, a "13" box, or a "0" box. Basically, we find the number of pairs of integers $(a,b)$ such that $a+b\le 33$ . This is $\dbinom{35}{2}=595$ Now, we must account for overcounting. Since $\gcd(13,20)=1$ , the first number that can be expressed as a multiple of 13 plus a multiple of 20 in more than one way is $13\cdot 20$ . $20\cdot 13=13\cdot 20$ A second collision would occur if we have $13m\cdot 20n$ , but since we only have 13 more powers to distribute on the left side, we cannot have $m,n>1$ .

Now, we have 13 more powers to distribute among both sides. The number of ways to do this, similarly to how we distributed the 33 powers initially, is $\dbinom{15}{2}=105$ The answer is $595-105=\boxed{490}$ .

The exponents of the non-zero terms are precisely the set $\{20a + 13b + 0c | a+b+c = 33\} = \{20a + 13b | a + b \leq 33\}.$ There are $34\cdot 35/2 = 595$ such pairs $(a,b)$ ; however, the values of $20a + 13b$ are not all unique.

Any pair $(a, b)$ with $b \geq 20$ generates the same exponent as the pair $(a+13, b-20)$ . Therefore we limit ourselves to the cases with $b < 20$ . SInce 13 and 20 are coprime, all values of $20a + 13b$ are now guaranteed to be unique.

Count therefore all pairs $(a,b)$ with $0 \leq a$ , $0 \leq b \leq 19$ , and $a + b \leq 33$ ; that gives $\sum_{b = 0}^19 (34-b) = 20\cdot 34 - \sum_{b=0}^19 b = 20\cdot 34 - \frac{34\cdot 35}2 = \boxed{490}.$