Le Monde puzzle [#1063]
A simple (summertime?!) arithmetic Le Monde mathematical puzzle
- 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?
- 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.)
August 10, 2018 at 12:17 pm
In the solution of the second part, 675 should not be in the list as it is 26^2-1 and 26 is an even number.
August 10, 2018 at 11:04 pm
Thanks, I forgot to add the odd constraint!
August 9, 2018 at 11:07 pm
[…] article was first published on R – Xi’an’s Og, and kindly contributed to […]
August 9, 2018 at 10:45 pm
Problem probably inspired by the Math exam this year for the Bachelor degree Section S , Math Speciality , Exercise No 4
August 10, 2018 at 11:05 pm
Great link, Jean-Louis, I will be back on that thread soon!!!
August 9, 2018 at 11:36 am
Seems the first question would be answered by the second one? As
is always powerful, then
and
are pair integers that satisfy the first question.
And we can show there are infinitely many: for any
which is a powerful integer,
will also be a powerful integer.