# Euler Problem 69

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.

``````n   Relatively Prime    φ(n)    n/φ(n)
2   1                   1   2
3   1,2                     2   1.5
4   1,3                     2   2
5   1,2,3,4             4   1.25
6   1,5                     2   3
7   1,2,3,4,5,6             6   1.1666...
8   1,3,5,7             4   2
9   1,2,4,5,7,8             6   1.5
10  1,3,7,9             4   2.5

``````

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.

``````

In [1]:

from sympy import primerange
P = 1
for p in primerange(1, 100):
P *= p
if P > 1000000:
break
print(P//p)

``````
``````

510510

``````

Explanation: n/φ(n) is equal to the product of p/(p-1) for all distinct prime divisors of n. We may assume that n has no repeated prime factors (i.e. is square free), because repeating a prime factor would increase the value of n without changing the value of n/φ(n). We may also assume that the prime factors are consecutive, because replacing one prime with a smaller prime would increase the product. So the answer is the largest product of consecutive primes (starting at 2) that is less than 1,000,000.