Euler’s Totient function, φ(n) [sometimes called the phi function], is used to determine the number of positive numbers less than or equal to n which are relatively prime to n. For example, as 1, 2, 4, 5, 7, and 8, are all less than nine and relatively prime to nine, φ(9)=6.
The number 1 is considered to be relatively prime to every positive number, so φ(1)=1.

Interestingly, φ(87109)=79180, and it can be seen that 87109 is a permutation of 79180.

Find the value of n, 1 <n < 10⁷, for which φ(n) is a permutation of n and the ratio n/φ(n) produces a minimum.

Analysis

We can reduce our search significantly by selecting prime pairs (p₁, p₂ ) and calculate n as n = p₁ x p₂. This allows the totient to be calculated as φ(n) = (p₁-1)(p₂-1) for n. Now, just find the minimum ratio n/φ(n) for those n and φ(n) that are permutations of one another.

The range of primes was selected by taking the square root of the upper bound, 10,000,000, which is about 3162 and taking ±25% for a range of primes from 2000 to 4000 (247 primes).

Solution

Runs < 3 seconds in Perl.

@p = split /s/,<DATA>;
($min_q, $min_n, $i, $max) = (2, 0, 1, 10_000_000);
 
for $p1 (@p) {
  for $p2 (@p[$i++ .. $#p]) { 
    $n = $p1 * $p2;
    last if $n > $max;
    $phi = ($p1-1) * ($p2-1);
    $q = $n / $phi;
    ($min_q, $min_n) = ($q, $n) if permutation($phi, $n) && $min_q > $q;
  } 
}
print "Answer to PE70 = $min_n";
 
sub permutation { join('', sort split //, @_[0]) eq join('', sort split //, @_[1]) }
__DATA__
2003 2011 2017 2027 2029 2039 2053 2063 2069 2081 2083 2087 2089 ... 3967 3989

Comments

The prime numbers from 2000 – 4000 were included at the end of the script and abbreviated here to save space.
We tried the same method for n, 1 <n < 10⁹, using the square root 31623 and the 25% bounds of 23,700 through 39,500. The result was 990326167 in about 50 seconds.

Euler’s Totient function, φ(n) [sometimes called the phi function], is used to determine the number of numbers less than n which are relatively prime to n. For example, as 1, 2, 4, 5, 7, and 8, are all less than nine and relatively prime to nine, φ(9)=6.

It can be seen that n=6 produces a maximum n/φ(n) for n ≤ 10.

Find the value of n ≤ 1,000,000 for which n/φ(n) is a maximum.

Analysis

We’re trying to find the smallest φ(n) and the largest n in order to keep n/φ(n) at a maximum. One method is to multiply as many consecutive primes as possible without exceeding the limit of 1,000,000.

Solution

Runs < 1 seconds in Python.

primes = [2,3,5,7,11,13,17,19,23,27,31,37,41]
limit=10**6
max = 1
for p in primes:
  if max * p > limit: break
  max *= p
print "Answer to PE69 = ", max

Comments

for 10⁷? 9,699,690 = 2 x 3 x 5 x 7 x 11 x 13 x 17 x 19, and following the same method for 10⁹ would be 223,092,870

The RSA encryption is based on the following procedure:

Generate two distinct primes p and q.
Compute n=pq and φ=(p-1)(q-1).
Find an integer e, 1<e<φ, such that gcd(e,φ)=1.

A message in this system is a number in the interval [0,n-1].
A text to be encrypted is then somehow converted to messages (numbers in the interval [0,n-1]).
To encrypt the text, for each message, m, c=m^e mod n is calculated.

To decrypt the text, the following procedure is needed: calculate d such that ed=1 mod φ, then for each encrypted message, c, calculate m=c^d mod n.

There exist values of e and m such that m^e mod n=m.
We call messages m for which m^e mod n=m unconcealed messages.

An issue when choosing e is that there should not be too many unconcealed messages.
For instance, let p=19 and q=37.
Then n=19*37=703 and φ=18*36=648.
If we choose e=181, then, although gcd(181,648)=1 it turns out that all possible messages
m (0≤m≤n-1) are unconcealed when calculating m^e mod n.
For any valid choice of e there exist some unconcealed messages.
It’s important that the number of unconcealed messages is at a minimum.

Choose p=1009 and q=3643.
Find the sum of all values of e, 1<e<φ(1009,3643) and gcd(e,φ)=1, so that the number of unconcealed messages for this value of e is at a minimum.

Analysis

At the heart of this problem is the simple formula: [gcd(e-1,p-1)+1][gcd(e-1,q-1)+1] as outlined in many texts such as Recent Advances in RCA Cryptography in the chapter Properties of the RSA cryptosystem. Specifically Lemma 5.1 on page 69 shows a nice proof of this equation.

So encoding this into a program is straightforward and simple. Just sum all values of e where gcd(e,phi) = 1 and gcd(e-1,q-1) and gcd(e-1,p-1) are at a minimum (equal 2).

Solution

Runs in under 18 seconds in Perl.

my ($p, $q, $sum) = (1009, 3643, 0); 
my $phi = ($p-1)*($q-1);
 
for (my $e=3; $e<$phi; $e+=4 ) {
  $sum += $e if gcd($e,$phi)==1 && gcd($e-1,$q-1)==2 && gcd($e-1,$p-1)==2;
}
 
print "Answer to PE 182 = $sum";

Comments

Runs in under 18 seconds.
Christner discovered the every 4 rule. See the gcd subroutine here.