## Fermat’s Riddle

Posted in Books, Kids, R with tags , , , , , , , , , , on October 16, 2020 by xi'an

·A Fermat-like riddle from the Riddler (with enough room to code on the margin)

An  arbitrary positive integer N is to be written as a difference of two distinct positive integers. What are the impossible cases and else can you provide a list of all distinct representations?

Since the problem amounts to finding a>b>0 such that

$N=a^2-b^2=(a-b)(a+b)$

both (a+b) and (a-b) are products of some of the prime factors in the decomposition of N and both terms must have the same parity for the average a to be an integer. This eliminates decompositions with a single prime factor 2 (and N=1). For other cases, the following R code (which I could not deposit on tio.run because of the packages R.utils!) returns a list

```library(R.utils)
library(numbers)
bitz<-function(i,m) #int2bits
c(rev(as.binary(i)),rep(0,m))[1:m]
ridl=function(n){
a=primeFactors(n)
if((n==1)|(sum(a==2)==1)){
print("impossible")}else{
m=length(a);g=NULL
for(i in 1:2^m){
b=bitz(i,m)
if(((d<-prod(a[!!b]))%%2==(e<-prod(a[!b]))%%2)&(d<e))
g=rbind(g,c(k<-(e+d)/2,l<-(e-d)/2))}
return(g[!duplicated(g[,1]-g[,2]),])}}
```

For instance,

```> ridl(1456)
[,1] [,2]
[1,]  365  363
[2,]  184  180
[3,]   95   87
[4,]   59   45
[5,]   40   12
[6,]   41   15
```

Checking for the most prolific N, up to 10⁶, I found that N=6720=2⁶·3·5·7 produces 20 different decompositions. And that N=887,040=2⁸·3²·5·7·11 leads to 84 distinct differences of squares.

## le compte est bon

Posted in Books, Kids, R with tags , , , , , , on July 22, 2020 by xi'an

The Riddler asks how to derive 24 from (1,2,3,8), with each number appearing once and all operations (x,+,/,-,^) allowed. This reminded me of a very old TV show on French TV, called Le compte est bon!, where players were given 5 or 6 numbers and supposed to find a given total within 60 ,seconds. Unsurprisingly there is an online solver for this game, as shown above, e.g., 24=(8+3+1)x2. But it proves unable to solve the puzzle when the input is 24 and (2,3,3,4), only using 2,3 and 4, since 24=2x3x4. Introducing powers as well, since exponentiation is allowed, leads to two solutions, (4-2)³x3=(4/2)³x3=(3²-3)x4=3/(2/4)³=24… Not fun!

I however rewrote an R code to check whether 24 was indeed a possibility allowed with such combinations but could not find an easy way to identify which combination was used, although a pedestrian version eventually worked! And exhibited the slightly less predictable 43/2x3=24!