The Riddler was about a number game, which meant starting a sum at 0 and adding i to the current sum at time i, almost a Fibonacci sequence but not exactly, since the sequence is simply
which is easy to code, is of period 200, and this code showed that 3, 28, 53, and 78 were the most frequent last two digits. Or all 3 modulo 25 digits.
A combinatorics puzzle from The Riddler about a Napoli solitaire where 4 x 10 cards numbered from 1 to 10 are shuffled and the game is lost when a number (1,2, or 3) is equal to its position modulo 3 (1,2 or 3). A simple R code shows that the probability of winning is around 0.00831:
N=40 for(t in 1:1e6)F=F+!sum(!(sample((1:N)%%10)-(1:N)%%3))
ChatGPT bends over backward to achieve this figure! Now, the exact probability can be found by combinatorics. While there are 40! ways of permuting the 40 cards, those missing the coincidences are
multiplied by 4!4!4!28! (which I initially forgot), resulting in 0.00831:
for(i in 0:4)for(j in 0:4)for(k in 0:4)
F=F+exp(lchoose(13,i)+lchoose(13,4-i)+3*lfactorial(4)+
lchoose(14,j)+lchoose(13-i,4-j)+lfactorial(28)+
lchoose(14-j,k)+lchoose(9+i,4-k)-lfactorial(40))
The weekly riddle from The Riddler is to find solutions to the Diophantine equation
c³-c=b²+4
(when b and c are positive integers). First, forget about ChatGPT since it states this is a Pell equation. With a wrong argument. Second, when running a basic R code, using as.double to handle larger integers, the only solution less than 10⁶ this code returned was
[1] 999799 999700015
with the first column being c and the second b. But this is not a correct solution!, as confirmed by Mathematica, which states there is no integer solution. This makes sense when looking at the unique real solution (in c) of the cubic
c³-c-(b²+4)=0
since the solution (using Cardano’s formula) involves
leaving the inverse of 27 as the only non-integer term in the expression when b is even… (The exact solution that this Diophantine equation has no solution is simpler: the lhs is a multiple of 3, while the rhs cannot be, as shown by looking at b(3).)
The New Year Riddle is about optimisation: starting with a single machine, between delivering one unit per machine – hour and delivering one new machine per machine every six days, what is the maximal number of units produced over 100 days?
Comparing the amounts produced by k machines after 6log2(k) days used to multiply the machines showed that 2¹⁵ -1 additional machines were first produced, to generate 7864320 items over the remaining 10 days. Which did not really require an R implementation (although I checked that intermediate solutions where only some of the machines were producing new machines were sub-optimal).
A simplistic puzzle from The Riddler when applying brute force:
A Tribonacci sequence is based on three entry integers a ≤ b ≤ c, and subsequent terms are the sum of the previous three. Among Tribonacci sequences containing 2023, which one achieves the smallest fourth term, a+b+c ?
The R code
![\sqrt[3]{\frac{b^2+4}{2}\pm\sqrt{\frac{(b^2+4)^2}{4}-\frac{1}{27}}}](https://s0.wp.com/latex.php?latex=%5Csqrt%5B3%5D%7B%5Cfrac%7Bb%5E2%2B4%7D%7B2%7D%5Cpm%5Csqrt%7B%5Cfrac%7B%28b%5E2%2B4%29%5E2%7D%7B4%7D-%5Cfrac%7B1%7D%7B27%7D%7D%7D&bg=000000&fg=B0B0B0&s=0&c=20201002)
tri<-function(a,b,e){
while(F<2023){
F=a+b+e;a=b;b=e;e=F}
return(F<2024)}
sol=NULL;m=674
for(a in 1:m)
for(b in a:m)
for(e in b:m)
if(tri(a,b,e)){
sol=rbind(sol,c(a,b,e))}
leads to (1,1,6) as the solution… Incidentally, this short exercise led me to finally look for a fix to entering vectors as arguments of functions requesting lists:
do.call("tri",as.list(sol[2023,]))
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