**R – Xi'an's Og**, and kindly contributed to R-bloggers]. (You can report issue about the content on this page here)

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**A**n 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?

**T**he integer x satisfies the identity

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

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…

Filed under: Books, Kids, R Tagged: arithmetics, competition, Le Monde, long division, mathematical puzzle, R

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