## Carmichael number, more or less

Posted in Books, Kids, Statistics with tags , , , , , , , on May 6, 2022 by xi'an

A quick-and-dirty R resolution of a riddle from The Riddler, namely to find a Carmichael number of the form abcabc:

```library(numbers)
for(i in 1:9)
for(j in 0:9)
for(k in 0:9){
x=i*100100+j*1010+k*101
if(!isPrime(x)){
p=primeFactors(x)
if((prod(apply(outer(p,p,F="=="),1,sum)%%2))&
(!max((x-1)%%(p-1))))break()}}
```

resulting into the number 101 101, since its prime factors are

```> primeFactors(101101)
[1]   7  11  13 101
```

and 6, 10, 12, and 100 are divisors of 101100:

```> primeFactors(101100)
[1] 2 2 3 5 5 337
```

## more games of life

Posted in Books, Kids, R with tags , , , , on May 5, 2020 by xi'an

Another puzzle in memoriam of John Conway in The Guardian:

Find the ten digit number, abcdefghij. Each of the digits is different, and

• a is divisible by 1
• ab is divisible by 2
• abc is divisible by 3
• abcd is divisible by 4
• abcde is divisible by 5
• abcdef is divisible by 6
• abcdefg is divisible by 7
• abcdefgh is divisible by 8
• abcdefghi is divisible by 9
• abcdefghij is divisible by 10

Which brute force R coding by checking over random permutations of (1,2,…,9) [since j=0] solves within seconds:

```while(0<1)
if (prod(!(x<-sum(10^{0:8}*sample(1:9)))%/%10^{7:0}%%2:9))break()
```

into x=3816547290. And slightly less brute force R coding even faster:

```while(0<1){
e=sample(c(2,6,8))#even
o=sample(c(1,3,7,9))#odd
if((!(o[1]+e[1]+o[2])%%3)&
(!(10*o[2]+e[2])%%4)&
(!(o[1]+e[1]+o[2]+e[2]+5+4)%%3)&
(!sum(10^{6:0}*c(o[1],e[1],o[2],e[2],5,4,o[3]))%%7)&
(!(10*o[3]+e[3])%%8)&
(!(sum(o)+sum(e))%%9)){
print(sum(10^{9:0}*c(o[1],e[1],o[2],e[2],4,5,o[3],e[3],o[4],0)));break()}}
```

## Le Monde puzzle [#913]

Posted in Books, Kids, Statistics, University life with tags , , , , , , , on June 12, 2015 by xi'an

An arithmetics Le Monde mathematical puzzle:

Find all bi-twin integers, namely positive integers such that adding 2 to any of their dividers returns a prime number.

An easy puzzle, once the R libraries on prime number decomposition can be found!, since it is straightforward to check for solutions. Unfortunately, I could not install the recent numbers package. So I used instead the schoolmath R package. Despite its possible bugs. But it seems to do the job for this problem:

```lem=NULL
for (t in 1:1e4)
if (prod(is.prim(prime.factor(t)+2)==1))
lem=c(lem,t)digin=function(n){
```

which returned all solutions, albeit in a lengthy fashion:

```> lem
[1] 1 3 5 9 11 15 17 25 27 29 33 41 45 51 55
[16] 59 71 75 81 85 87 99 101 107 121 123 125 135 137 145
[31] 149 153 165 177 179 187 191 197 205 213 225 227 239 243 255
[46] 261 269 275 281 289 295 297 303 311 319 321 347 355 363 369
[61] 375 405 411 419 425 431 435 447 451 459 461 493 495 505 521
[76] 531 535 537 561 569 573 591 599 605 615 617 625 639 641 649
[91] 659 675 681 685 697 717 725 729 745 765 781 783 807 809 821
[106] 825 827 841 843 857 867 881 885 891 895 909 933 935 955 957
[121] 963 985 1003 1019 1025 1031 ...
```