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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()}}
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