What Would You Call A Non-base 10 Decimal?

Let us investigate $k = 4$ . We find that the decimal is equal to $\frac{1}{4} + \frac{3}{16} + \frac{3}{64} + \frac{1}{256} + \frac{3}{1024} + \frac{3}{4096}$ . It may not be obvious at first, but this series is geometric. We can take advantage of the fact that the decimal is repeating to group terms. Indeed, by grouping each consecutive three terms, we find the sum $\frac{16 + 3(4) + 3}{64} + \frac{16 + 3(4) + 3}{256}...$ . Using the geometric series formula with common ratio $\frac{1}{4}$ , we find $\frac{\frac{31}{64}}{1 - \frac{1}{64}} = \frac{31}{63}$ . What a handy closed form! We should be able to generalize this. Namely, $a_k = \frac{k^2 + 3k + 3}{k^3 - 1}$ for $k \geq 4$ . Now, this is still quite difficult to evaluate, so let's try some more messing around.

We can factorize the denominator to get $\frac{k^2 + 3k + 3}{(k-1)(k^2 + k + 1)}$ . So if we investigate the $\frac{1}{k-1}$ part, when we take the product from $4$ to $99$ we will get $\frac{2}{98!}$ . So let's take out that part to simplify our expression.

We are left to evaluate $\frac{2}{98!} \displaystyle \prod_{k = 4}^{99} \frac{k^2 + 3k + 3}{k^2 + k + 1}$ . Using the principle of hopeful elegance (a useful tool which says that you should always be actively looking for elegant ways to solve problems, "hoping" that one will show up as it often does) or through some very solid algebraic intuition, this product telescopes. Namely, in the product $\frac{k^2 + 3k + 3}{k^2 + k + 1} \times \frac{(k+1)^3 + 3(k+1) + 3}{(k+1)^2 + (k + 1) + 1} \times \frac{(k+2)^2 + 3(k+2) + 3}{(k+2)^2 + (k+2) + 1} \cdots$ , the numerator of the previous term cancels out with the denominator of the next term (since $(n+1)^2 + (n+1) + 1 = n^2 + 2n + 1 + n + 1 + 1 = n^2 + 3n + 3$ )

Thus, if we take this product from $k = 4$ to $99$ , we will be left with the denominator of $k=4$ and the numerator of $k = 99$ . So our product comes out to $\frac{2}{98!} \times \frac{99^2 + 3*99 + 3}{4^2 + 3*4 + 3} = \frac{962}{98!}$ .

Finally, for our answer, $\frac{962}{98} = \frac{481}{49}$ , so $p+q = 530$ .

We can write the repeating decimal $0.133..._{(k)}$ as an infinite series:

$0.133..._{(k)} = \sum_{i=1}^\infty \frac{1}{k^{3i+1}} + \frac{3}{k^{3i+2}} + \frac{3}{k^{3i+3}} = \frac{-(3 + 3*k + k^2)}{1 - k^3}_{(10)}$

$\prod_{k=4}^{q} \frac{-(3 + 3*k + k^2)}{1 - k^3} = \frac{2(q^2 + 3q + 3)}{21(q-1)!}$

Plugging in $q = 99$ , we get a result of

$\frac{962}{98!} = \frac{m}{n!}$

therefore $p = 481$ , $q = 49$ , $p + q = 530$

Lets find a general expression for $a_k,\: k\geq 4$ . Remember that in base $k$ , a shift by one digit to the right equals a multiplication by $k^{-1}$ : $\begin{aligned} a_k&=\left(0.\overline{\red{1}\green{3}\blue{3}}\right)_k=\left(\red{1}\cdot k^{-1}+\green{3}\cdot k^{-2}+\blue{3}\cdot k^{-3}\right)\cdot\sum_{i=0}^\infty (k^{-3})^i&&\left|\text{Geom. Series,}\right.\quad\left|k^{-3}\right|\leq \frac{1}{4^3}<1\\ &=\frac{ k^{-1}+3k^{-2}+3k^{-3} }{1-k^{-3}}\cdot\frac{k^4}{k^4}=\frac{ k^3+3k^2+3k\pm 1 }{ k(k^3-1) }&&\left|(k+1)^3=k^3+3k^2+3k+1\right.\\ &=\frac{1}{k}\cdot\frac{(k+1)^3-1}{k^3-1}\\ \end{aligned}$

Put all that hard work to use simplifying the product. Notice the second factor is a telescopic product: $\begin{aligned} \prod_{k=4}^{99}a_k&=\left( \prod_{k=4}^{99}\frac{1}{k} \right)\cdot\prod_{k=4}^{99}\frac{(k+1)^3-1}{k^3-1}=\frac{3!}{99!}\cdot\frac{100^3-1}{4^3-1}=\frac{ \cancel{6}^2\cdot \cancel{99}\cdot\cancel{21}\cdot 481 }{ \cancel{99}\cdot 98!\cdot \cancel{63} }=\frac{962}{98!}=:\frac{m}{n!} \end{aligned}$

As $962\mod 98=80\neq0$ , we can not further reduce the factorial:

$\begin{aligned} \frac{m}{n}&=\frac{962}{98}=\frac{481}{49}=\frac{p}{q}&\Rightarrow&&p+q&=\boxed{530} \end{aligned}$

Our $a_{k}$ is defined as $k^{-1} + 3*k^{-2} + 3*k^{-3} + k^{-4} + ...$

We can reorder the terms of the sum into \((k^{-1} + k^{-4} + k^{-7} + ...) + 3*(k^{-2} + k^{-5} + k^{-8} + ... ) + 3*(k^{-3} + k^{-6} + k^{-9} + ...)

= (1/k)/(1 - 1/k^{3}) + 3*(1/k^{2})/(1 - 1/k^{3}) + 3*(1/k^{3})/(1 - 1/k^{3})

= k^{2}/(k^{3} - 1) + 3{k}/(k^{3} - 1) + 3/(k^{3} - 1) = \frac{k^2 + 3k + 3}{k^3 - 1}\)

Note that $a_{k + 1} = (k^2 + 2k + 1 + 3k + 3 + 3)/(k^3 + 3k^2 + 3k + 1 - 1)$

$a_{k+1} = (k^2 + 5k + 7)/(k^3 + 3k^2 + 3k) = (1/k)*(k^2 + 5k + 7)/(k^2 + 3k + 3)$

Thus, when calculating the product $a_{k}*a_{k+1}$ and applying it from 4 to 99, the only things that will remain in the product are the first denominator and the last numerator divided by the product of the numbers

$= [(99^2 + 3*99+ 3)*(1/63)]/(4*5*6*...*98)$

$= [(10000 - 200 + 1 + 300 - 3 + 3)/63]*6/98! = [60606/63]/98! = 962/98!$

Then, $m/n = 962/98 = 481/49$ , thus $p = 481, q = 49, p + q = \boxed{530}$