Archive for digits

Le Monde puzzle [#1111]

Posted in R, Statistics with tags , , , , , , on September 18, 2019 by xi'an

Another low-key arithmetic problem as Le Monde current mathematical puzzle:

Notice that there are 10 numbers less than, and prime with 11, 100 less than and prime with 101, 1000 less than, and prime with 1111? What is the smallest integer N such that the cardinal of the set of M<N prime with N is 10⁴? What is the smallest multiple of 1111 using only different digits? And what is the largest?

As it is indeed a case for brute force resolution as in the following R code

library(numbers)
homanycoprim=function(n){
  many=1
  for (i in 2:(n-1)) many=many+coprime(i,n)
  return(many)}

smallest=function(){
  n=1e4
  many=homanycoprim(n)
  while (many!=1e4) many=homanycoprim(n<-n+1)
  return(n)}

which returns n=10291 as the smallest value of N.  For the other two questions, the usual integer2digit function is necessary

smalldiff=function(){
  n=1111;mul=1
  while (mul<1e6) {
    x=as.numeric(strsplit(as.character(n*mul), "")[[1]])
    while (sum(duplicated(x))!=0){
     mul=mul+1
     x=as.numeric(strsplit(as.character(n*mul), "")[[1]])}
  print(n*mul);mul=mul+1}}

leading to 241,087 as the smallest and 9,875,612,340 as the largest (with 10 digits).

Le Monde puzzle [#1110]

Posted in Books, Kids, R, Travel with tags , , , , , , , , , on September 16, 2019 by xi'an

A low-key sorting problem as Le Monde current mathematical puzzle:

If the numbers from 1 to 67 are randomly permuted and if the sorting algorithm consists in picking a number i with a position higher than its rank i and moving it at the correct i-th position, what is the maximal number of steps to sort this set of integers when the selected integer is chosen optimaly?

As the question does not clearly state what happens to the number j that stood in the i-th position, I made the assumption that the entire sequence from position i to position n is moved one position upwards (rather than having i and j exchanged). In which case my intuition was that moving the smallest moveable number was optimal, resulting in the following R code

sorite<-function(permu){ n=length(permu) p=0 while(max(abs(permu-(1:n)))>0){
    j=min(permu[permu<(1:n)])
    p=p+1
    permu=unique(c(permu[(1:n)<j],j,permu[j:n]))} 
  return(p)}

which takes at most n-1 steps to reorder the sequence. I however checked this insertion sort was indeed the case through a recursive function

resorite<-function(permu){ n=length(permu);p=0 while(max(abs(permu-(1:n)))>0){
    j=cand=permu[permu<(1:n)]
    if (length(cand)==1){
      p=p+1
      permu=unique(c(permu[(1:n)<j],j,permu[j:n]))
    }else{
      sol=n^3
      for (i in cand){
        qermu=unique(c(permu[(1:n)<i],i,permu[i:n]))
        rol=resorite(qermu)
        if (rol<sol)sol=rol}
      p=p+1+sol;break()}} 
  return(p)}

which did confirm my intuition.

Le Monde puzzle [#1109]

Posted in Kids, R with tags , , , , , on September 4, 2019 by xi'an

A digital problem as Le Monde current mathematical puzzle:

Noble numbers are such that they only involve different digits and are multiple of all their digits. What is the largest noble number?

Hmmmm…. Brute force? Since the maximal number of digits is 10, one may as well try:

k=soz=9
for (t in 1:1e3){
sol=1
while (sol<10^(k-1)){ 
 u=sample(0:8,k);i=digit2int(u) 
 if (max(i%%u[u>0])==0) soz=max(soz,sol<-i)}}

which returns 9875643120… (I made the conscious choice to exclude zero from the dividers. Which was not a choice spelled out in the original question.)

Le Monde puzzle [#818]

Posted in Books, Kids, R with tags , , , , on May 1, 2013 by xi'an

The current puzzle is as follows:

Define the symmetric of an integer as the integer obtained by inverting the order of its digits, eg 4321 is the symmetric of 1234. What are the numbers for which the square is equal to the symmetric of the square of the symmetric?

I first consulted stackexchange to find a convenient R function to create the symmetric:

int2digit=function(x){
as.numeric(sapply(sequence(nchar(x)),
  function(y) substr(x, y, y)))}

digit2int=function(a){
as.numeric(paste(a,collapse=""))}

flip=function(x){
  digit2int(rev(int2digit(x)))}

and then found that all integers but the multiples of 10 are symmetric! only some integers involving 1,2,3 and sometimes zero would work:

> for (i in 1:1000){
+   if (i^2==flip(flip(i)^2)) print(i)}
[1] 1
[1] 2
[1] 3
[1] 11
[1] 12
[1] 13
[1] 21
[1] 22
[1] 31
[1] 101
[1] 102
[1] 103
[1] 111
[1] 112
[1] 113
[1] 121
[1] 122
[1] 201
[1] 202
[1] 211
[1] 212
[1] 221
[1] 301
[1] 311

If I am not (again) confused, the symmetric integers would be those (a) not ending with zero and (b) only involving digits whose products are all less than 10.

Anomalies in the Iranian election

Posted in Statistics with tags , , on June 17, 2009 by xi'an

While the results of the recent Iranian presidential election are currently severely contested, with accusations of fraud and manipulations, and with an level of protest unheard of in Iran, I had not so far seen a statistical analysis of the votes. This is over: Boudewijn Roukema, a cosmologist with the University of Toruń (Poland), has produced an analysis of the figures published by the Iranian Ministry of the Interior, based on Benford’s Law for the repartition of the first digit i in decimal representations of real numbers, which should be

f(i) \propto \log_{10}(1+\frac{1}{i})

for the proportion of votes for a candidate among the four in competition. Roukema exhibits a very unlikely discrepancy on Mehdi Karoubi’s votes, with an extremely high occurence of the digit 7. There is also a discrepancy for Mahmoud Ahmadinejad’s frequencies of 1’s and 2’s that is harder to detect because of the higher frequency of votes for this candidate in the Iranian Ministry of the Interior data. But looking at the most populous districts, Roukema concludes that several million votes could have been added to Ahmadinejad’s votes in those areas, if Benford’s Law holds…

I find this analysis produced merely five days after the election quite astounding, even though the validity of applying Benford’s Law in those circumstances needs more backup…