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2 answers

Doug's computations are correct. These values are small enough you could do them by hand, but let me explain two things that make it easy to figure out Euler Φ values.

Φ(ab) = Φ(a)*Φ(b) for relatively prime a & b
Φ(p^k) = p^{k-1} * (p - 1) for a prime p (so Φ(p) = p - 1)

So Φ(40) = Φ(8)*Φ(5) = 4*4 = 16 (since Φ(2^3) = 4*Φ(2) = 4)
Φ(41) = 40 (since 41 is prime)
Φ(42) = Φ(2)*Φ(3)*Φ(7) = 1*2*6 = 12

2007-03-01 01:17:07 · answer #1 · answered by brashion 5 · 2 0

I'm guessing that you're talking about Eulers phi function Φ(n) defined to be equal to the number of integers between 0 and n-1 which are relatively prime to n. If so
Φ(40) = 16
Φ(41) = 40
Φ(42) = 12

HTH ☺

Doug

2007-02-28 22:39:16 · answer #2 · answered by doug_donaghue 7 · 2 0

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