Plentiful Pile of Polynomials

Our polynomial is $P(x)=a_0+a_1x+a_2x^2+\dots$ such that for each $a_i$ we have $0 \leq a_i \leq 24$ . We are looking for polynomials such that $P(5)=a_0+a_1 5+a_2 25+\dots=1200$ . This problem is equivalent to finding the number of unordered partitions of 1200 into parts of size $5^k$ for some non-negative integer $k$ such that each part occurs no more than 24 times. We proceed with generating functions.

The Generating function for parts of size $n$ that don't appear more than 24 times is $1+x^n+x^{2n}+\dots+x^{24n}$ , which we can rewrite as

$\frac{x^{25n}-1}{x^n-1}$

Thus, we are looking for the coefficient of $x^{1200}$ in

$\frac{x^{25\times 1}-1}{x^1-1}\frac{x^{25\times 5}-1}{x^5-1}\frac{x^{25\times 25}-1}{x^{25}-1}\frac{x^{25\times 125}-1}{x^{125}-1}\cdots$

Notice that the numerator of the $n^{th}$ term cancels with the denominator of the $(n+2)^{th}$ term, so we are left with

$\frac{x^{25\times 5^k}-1}{x^1-1}\frac{x^{25\times 5\times 5^k}-1}{x^5-1}$

as $k\to\infty$ .

This is the same as

$(1+x+x^2+\dots)(1+x^5+x^{25}+\dots)$

Since we want to find the coefficient of $x^{1200}$ in the previous expression, our problem now is to find the number of ordered pairs $(a,b)$ such that $a+5b=1200$ , which is far easier to work with.

Since each choice for $b$ uniquely determines a valid $a$ and we can choose $b$ on $[0,240]$ , we have 241 solutions, and as a result, 241 distinct polynomials $P(x)$ . So our answer is $\boxed{241}$

Since $5^5 \gt 1200$ we only need to consider polynomials of degree at most $4$ .

$P(x) = a_4x^4 + a_3x^3 + a_2x^2 + a_1x + a_0,$ and then set $x=5$ and rearrange.

We can write $P(5) = 1200$ as $\begin{aligned} 1200 &= (a_4 25^2 + a_2 25 + a_0) + 5(a_3 25 + a_1) \\ &= c + 5d \end{aligned}$

where $c$ represented in base $25$ has digits $a_4a_2a_0$ , and $d$ has digits $a_3a_1$ .

So there is a bijection between pairs $\{c,d\}$ and polynomials $P$ with coefficients that are valid digits in base $25$ .

Since $d$ can range from $0$ to $240$ , and for each $d$ there is exactly one $c$ such that $1200 = c + 5d$ , there are $\boxed{241}$ possible polynomials.

Write $P(x)$ as

$P(x)= \sum_{k=0}^\infty (a_k+5b_k)x^k$

for some sequences $a_k,b_k \in \{0,1,2,3,4\}$ (only finitely many of which are non-zero). Then $P(5)=1200$ means that $a_0=0$ and

$240= \sum_{k=1}^\infty a_k 5^{k-1} + \sum_{k=0}^\infty b_k 5^k$

Now notice that for each $N \in \{0,1,...,240\}$ the equations:

$\sum_{k=1}^\infty a_k 5^{k-1} =N$

$\sum_{k=0}^\infty b_k 5^k =240-N$

each have one unique solution. (Just consider the base- $5$ representations of $N$ and $240-N$ .) Hence, the total number of polynomials is $|\{0,...,240\}|=241$ .

When speaking of a polynomial $P$ , let's denote its coefficients with $p_i$ . Let's denote the number of polynomials $P(x)$ such that $P(5) = a$ with $C(a)$ . Obviously, for $0 \le a < 5$ , $C(a) = 1$ (the only such polynomial is $P(x) = a$ ).

For $a > 5$ , $5 \mid (P(5) - p_0)$ , therefore $P(5) \equiv p_0 \mod{5}$ . Hence, $C(a) = \sum_{0 \le i \le 24\ where\ i \equiv a \mod{5}} C(\frac{a - i}{5})$ (such polynomials have form $P(x) = i + x * P^\prime(x)$ ).

Using this, we can easily calculate pretty much any value of $C$ . Here are the ones we need:

$C(5) = C(\frac{5 - 0}{5}) + C(\frac{5 - 5}{5}) = C(0) + C(1) = 2$

$C(6 \leq a \leq 9) = C(0) + C(1) = 2$

$C(43) = C(44) = C(8) + C(7) + C(6) + C(5) + C(4) = 9$

$C(45) = C(46) = C(47) = C(9) + C(8) + C(7) + C(6) + C(5) = 10$

$C(236 \leq a \leq 239) = C(47) + C(46) + C(45) + C(44) + C(43) = 49$

$C(240) = C(48) + C(47) + C(46) + C(45) + C(44) = 49$

$C(1200) = C(240) + C(239) + C(238) + C(237) + C(236) = \boxed{241}$

We know that maximum degree of polynomial can be 4 and minimum 3...Let the polynomial be ax^4 + bx^3 + cx^2 + dx+e.(e is always 0,5,10,15 or 20) With this in the mind we do it for two cases.. first one when degree=4 that is. a=1 then 125b+25c+5d+e=575(for x=5)
now keeping b=0,1,2,3,4 we get 25c+5d+e=575,450,325,200,75 respectively.
now taking each case individually . we get total (25+25+25+25+16) =116 solutions.
Which can be obtained easily. As we know that maximum value of 5d+e can be 125 only as it has to be a multiple of 25 (so it cant be 130 to 140) taking maximum value of c=23 then 5d+e=0 , similarly for (c=22 to 18) 5d+e varies from 125 to 25. for 5d+e=125(for d,e=(24,5),(23,10),(22,15),(21.20) . for 5d+e=100(d,e=(20,0),19,5),(18,10),(17,15),(16,20) . similarly for 5d+e=(75,50,25) gives 5 solutions each.
which in total giving (5into4 +4 +1=25) solutions.. similarly for b=1,2,3,4 25c+5d+e=450,325,200 gives 25 solutions each. and for last case 75 gives16 solutions . as 5d+e=75,50,25 only.. each giving 5 solutions like before. and 1 more comes when 5d+e=0.totaling 5into3+1=16 solutions so in total (25+25+25+25+16) =116 solutions.

now for case two we keep a=0 and degree 3 then minimum value of b=4 as 25c+5d+e has maximum value 744.. so for b=4 25c+5d+e=700 for c=23 5d+e=125 has 4 solutions and c=22 5d+e=100 gives 5 solutions.. and now for b=5 25c+5d+e=575 same as case 1 which gives again 116 solutions. so in all we have 116into2 +5+4=241 solutions..

First observe that the degree of $P(x)$ is either $3$ or $4$ . Case 1: Let $P(x)=ax^3+bx^2+cx+d$ where $a,b,c,d$ lie in the given range, and $a \neq 0$ .

$P(5)=1200=25 \cdot 48=125a+25b+5c+d$

We note that since LHS is a multiple of $25$ so must be LHS. Thus, it follows that $5c+d$ must be a multiple of $25$ . $0 \leq 5c+d \leq 144$ $\Rightarrow 5c+d \in (0,25,50,75,100,125)$ We further note that $d$ is a multiple of $5$ . Let $d=5k$ . Then, $0 \leq k \leq 4$ . $c+k \in (0,5,10,15,20,25)$ Now, its easy to spot that we have $5$ solutions to all except when $c+k=0$ , and $c+k=25$ . For $c+k=0$ there is a just $1$ solution, and $4$ solutions when $c+k=25$ , since $c < 25 \Rightarrow k \neq 0$ in this particular case.

We now establish that $25(5a+b) \in (1200,1175,1150,1125,1100,1075)$ $(5a+b) \in (48,47,46,45,44,43)$ Via modular arithmetic or simple reasoning, we have that $b \in (5t+3,5t+2,5t+1,5t,5t+4,5t+3)$ for the the above cases, respectively. Hence, $(a+t) \in (9,9,9,9,8,8)$ . Clearly , $0 \leq t \leq 4$ satisfies all the cases. $5$ solutions exist in each case, which can easily be checked. Thus number of solutions in this case: $=5\cdot 1 + 4 \cdot 5^2 + 4 \cdot 5=125$ .

Case 2: $P(x)=ax^4+bx^3+cx^2+dx +e$ $P(5)=25 \cdot 48 = 625a + 125b+25c+5d+e=1200$ Note that $625 a \leq 1200 \Rightarrow a=1$ $125b+25c+5d+e=575$ An analogous argument as in the previous case, tell us that we have $5d+e \in (0,25,50,75,100,125)$ .We already know the number of solutions for each case here.

And then we have : $5b+c \in (23,22,21,20,19,18)$ - $5b+c=23,22,21,20$ ---> $0 \leq b \leq 4$ ----> $5$ solutions for each. - $5b+c=19,18$ ---> $0 \leq b \leq 3$ ----> 4 solutions for each.

Total solutions in this case : $=5 \cdot 1 +3 (\cdot 5^2) + 4\cdot 5+5\cdot 4=116$

TOTAL number of solutions : $= \boxed{241}$

Consider the polynomials $f(x)=1+x+x^2+...+x^{24}$ and $g(x)=1+x+x^2+...+x^5$ .

The problem is asking us to find the coefficient of $x^{1200}$ in the expansion $f(x)f\left(x^5\right)f\left(x^{25}\right)f\left(x^{125}\right)...$

Note that we have the factorization $f(x)=g(x)g(x^5)$ . Thus, we want to compute the coefficent of $x^{1200}$ in $\left(g(x)g\left(x^5\right)g\left(x^{25}\right)...\right)\left(g\left(x^5\right)g\left(x^{25}\right)...\right).$

However, since every number has a unique base five expression, we have $g(x)g\left(x^5\right)g\left(x^{25}\right)... = 1 + x+x^2+x^3+... = \dfrac{1}{1-x}.$

Substituting in to the earlier expression, we want the coefficient of $x^{1200}$ in $\dfrac{1}{1-x}\cdot\dfrac{1}{1-x^5}$ which upon expansion is clearly $\left[\dfrac{1200}{5}\right]+1=241.$