Archive for arithmetics

another attempt at code golf

Posted in Books, Kids, R with tags , , , on June 12, 2019 by xi'an

I had another lazy weekend go at code golf, trying to code in the most condensed way the following task. Provided with a square matrix A of positive integers, keep iterating the steps

  • take the highest square 𝑥² in A.
  • find the smallest adjacent neighbour 𝑛
  • replace with x and n with nx

until no square is left (with neighbour defined as either horizontally or vertically and without wrapping around). While I managed a 217 bytes solution, compared with Robin’s 179b improvement, which remains surprising readable!, the puzzle offers two further questions:

  1. is there a non-iterative way to find the final matrix B?
  2. the puzzle assumes that A satisfies that at each step, the highest square and the smallest neighbour n will be unique, and that the sequence will not repeat forever. Is there a fool-proof way to check this is the case?

 

an attempt at code golf

Posted in Kids, R with tags , , , , , , , , , on May 15, 2019 by xi'an

Having discovered codegolf on Stack Exchange a few weeks ago, I spotted a few interesting puzzles since then but only got the opportunity at a try over a quiet and rainy weekend (and Robin being on vacation)! The challenge was to write an R code for deciding whether or not a given integer n is congruent or not, when congruent means that it is the surface of a rectangle triangle with all three sides rational. The question included a pointer to the Birch and Swinnerton-Dyer conjecture as a mean to check congruence although the real solution was provided by Tunnell’s Theorem, which states that n is congruent if and only if the number of integer solutions to 2x²+y²+8z²=n is twice as much as the number of integer solutions to 2x²+y²+32z²=n if n is odd and  the number of integer solutions to 8x²+y²+16z²=n is twice as much as the number of integer solutions to 8x²+y²+64z²=n if n is even. Although this is only true for squared-free integers. (I actually spent more time on figuring out the exact wording of the theorem than on optimising the R code!)

My original solution

p=function(n){
 for (i in(n:2)^2)if(n%%i<1)n=n/i
 if(n%%2){d=8;f=2;g=16}else{d=2;f=1;g=8}
 A=0;b=(-n:n)^2
 for(x in d*b)for(y in x+f*b)for(z in g*b)
  A=A+(y+z==n)-2*(y+4*z==n)
 A==0}

was quite naïve, as shown by the subsequent improvements by senior players, like the final (?) version of Guiseppe:

function(n){b=(-n:n)^2
for(i in b[b>0])n=n/i^(!n%%i)
P=2^(n%%2)
o=outer
!sum(!o(y<-o(8/P*b,2*b,"+")/P-n,z<-16/P*b,"+"),-2*!o(y,4*z,"+"))}

exhibiting a load of code golf tricks, from using an anonymous function to renaming functions with a single letter, to switching from integers to booleans and back with the exclamation mark.

Le Monde puzzle [#1083]

Posted in Books, Kids, R, Travel with tags , , , , , , on February 7, 2019 by xi'an

A Le Monde mathematical puzzle that seems hard to solve without the backup of a computer (and just simple enough to code on a flight to Montpellier):

Given the number N=2,019, find a decomposition of N as a sum of non-trivial powers of integers such that (a) the number of integers in the sum is maximal or (b) all powers are equal to 4.  Is it possible to write N as a sum of two powers?

It is straightforward to identify all possible terms in these sums by listing all powers of integers less than N

pool=(1:trunc(sqrt(2019)))^2
for (pow in 3:11)
  pool=unique(c(pool,(2:trunc(2019^(1/pow)))^pow))

which leads to 57 distinct powers. Sampling at random from this collection at random produces a sum of 21 perfect powers:

 1+4+8+9+16+25+27+32+36+49+64+81+100+121+125+128+144+169+196+243+441

But looking at the 22 smallest numbers in the pool of powers leads to 2019, which is a sure answer. Restricting the terms to powers of 4 leads to the sequence

1⁴+2⁴+3⁴+5⁴+6⁴ = 2019

And starting from the pools of all possible powers in a decomposition of 2019 as the sum of two powers shows this is impossible.

Le Monde puzzle [#1075]

Posted in Books, Kids, R with tags , , , , on December 12, 2018 by xi'an

A new Le Monde mathematical puzzle in the digit category:

Find the largest number such that each of its internal digits is strictly less than the average of its two neighbours. Same question when all digits differ.

For instance, n=96433469 is such a number. When trying pure brute force (with the usual integer2digits function!)

le=solz=3
while (length(solz)>0){
 solz=NULL
 for (i in (10^(le+1)-1):(9*10^le+9)){
  x=as.numeric(strsplit(as.character(i), "")[[1]])
 if (min(x[-c(1,le+1)]<(x[-c(1,2)]+x[-c(le,le+1)])/2)==1){ print(i);solz=c(solz,i); break()}}
 le=le+1}

this is actually the largest number returned by the R code. There is no solution with 9 digits. Adding an extra condition

le=solz=3
while (length(solz)>0){
 solz=NULL
 for (i in (10^(le+1)-1):(9*10^le+9)){
  x=as.numeric(strsplit(as.character(i), "")[[1]])
 if ((min(x[-c(1,le+1)]<(x[-c(1,2)]+x[-c(le,le+1)])/2)==1)&
    (length(unique(x))==le+1)){ print(i);solz=c(solz,i); break()}}
 le=le+1}

produces n=9520148 (seven digits) as the largest possible integer.

Le Monde puzzle [#1063]

Posted in Books, Kids, R with tags , , , , , , on August 9, 2018 by xi'an

lemondapariA simple (summertime?!) arithmetic Le Monde mathematical puzzle

  1. A “powerful integer” is such that all its prime divisors are at least with multiplicity 2. Are there two powerful integers in a row, i.e. such that both n and n+1 are powerful?
  2.  Are there odd integers n such that n² – 1 is a powerful integer ?

The first question can be solved by brute force.  Here is a R code that leads to the solution:

isperfz <- function(n){ 
  divz=primeFactors(n) 
  facz=unique(divz) 
  ordz=rep(0,length(facz)) 
  for (i in 1:length(facz)) 
    ordz[i]=sum(divz==facz[i]) 
  return(min(ordz)>1)}

lesperf=NULL
for (t in 4:1e5)
if (isperfz(t)) lesperf=c(lesperf,t)
twinz=lesperf[diff(lesperf)==1]

with solutions 8, 288, 675, 9800, 12167.

The second puzzle means rerunning the code only on integers n²-1…

[1] 8
[1] 288
[1] 675
[1] 9800
[1] 235224
[1] 332928
[1] 1825200
[1] 11309768

except that I cannot exceed n²=10⁸. (The Le Monde puzzles will now stop for a month, just like about everything in France!, and then a new challenge will take place. Stay tuned.)

seven digit addition

Posted in Kids, R with tags , , , on July 6, 2018 by xi'an

Another quick riddle from the riddler: solve the equation

EXMREEK + EHKREKK = ?K?H?X?E

which involves every digit between 0 and 9. While the puzzle can be unravelled by considering first E and K, which must be equal to 6 and 3, a simple R code also leads to the conclusion

isok <- function(a,b){
 s=as.numeric(unlist(strsplit(as.character(sum(10^(6:0)*a)+
   sum(10^(6:0)*b)),"")))
 if (length(s)==7){ goal=FALSE}else{
   goal=(length(unique(c(a,b,s)))==10)&(a[2]==s[6])&
         (s[8]==a[6])&(s[2]==a[7])&(b[2]==s[4])}
 return(goal)}

pasok <- function(T=1e3){ 
for (t in 1:T){ 
  a[1]=a[5]=a[6]=6;a[7]=3 
  digs=sample(c(0:2,4,5,7:9),4) 
  a[2:4]=digs[1:3] b[1]=a[1];b[2]=digs[4];
  b[3]=a[7];b[4]=a[4];b[5]=a[1];b[6:7]=a[7] 
  if (isok(a=a,b=b)) 
     print(rbind(a,b))}} 

> pasok()
  [,1] [,2] [,3] [,4] [,5] [,6] [,7]
a    6    2    4    7    6    6    3
b    6    8    3    7    6    3    3

which sum is 13085296.

Le Monde puzzle [#1018]

Posted in Books, Kids, R with tags , , , , , on August 29, 2017 by xi'an

An arithmetic Le Monde mathematical puzzle (that first did not seem to involve R programming because of the large number of digits in the quantity involved):

An integer x with less than 100 digits is such that adding the digit 1 on both sides of x produces the integer 99x.  What are the last nine digits of x? And what are the possible numbers of digits of x?

The integer x satisfies the identity

10^{\omega+2}+10x+1=99x

where ω is the number of digits of x. This amounts to

10….01 = 89 x,

where there are ω zeros. Working with long integers in R could bring an immediate solution, but I went for a pedestrian version, handling each digit at a time and starting from the final one which is necessarily 9:

#multiply by 9
rap=0;row=NULL
for (i in length(x):1){
prud=rap+x[i]*9
row=c(prud%%10,row)
rap=prud%/%10}
row=c(rap,row)
#multiply by 80
rep=raw=0
for (i in length(x):1){
prud=rep+x[i]*8
raw=c(prud%%10,raw)
rep=prud%/%10}
#find next digit
y=(row[1]+raw[1]+(length(x)>1))%%10

returning

7 9 7 7 5 2 8 0 9

as the (only) last digits of x. The same code can be exploited to check that the complete multiplication produces a number of the form 10….01, hence to deduce that the length of x is either 21 or 65, with solutions

[1] 1 1 2 3 5 9 5 5 0 5 6 1 7 9 7 7 5 2 8 0 9
[1] 1 1 2 3 5 9 5 5 0 5 6 1 7 9 7 7 5 2 8 0 8 9 8 8 7 6 4 0 4 4 9 4 3 8 2 0 2 2
[39] 4 7 1 9 1 0 1 1 2 3 5 9 5 5 0 5 6 1 7 9 7 7 5 2 8 0 9

The maths question behind is to figure out the powers k of 10 such that

10^k\equiv -1 \text{ mod } (89)

For instance, 10²≡11 mod (89) and 11¹¹≡88 mod (89) leads to the first solution ω=21. And then, since 10⁴⁴≡1 mod (89), ω=21+44=65 is another solution…