Totatives of 3600

To compute the number of totatives of n, you can use the formula:

$\phi(n) = n \prod\limits_{p|n} (1-\frac{1}{p})$

in which the product is over the distinct prime numbers dividing n.

The prime factorisation of 3600 is $2^{4} \times 3^{2} \times 5^{2}$ so the distinct primes dividing 3600 are 2, 3 and 5. Therefore:

$\phi(3600) = n \prod\limits_{p|3600} (1-\frac{1}{p})$ = $3600 \times (1-\frac{1}{2}) \times (1-\frac{1}{3}) \times (1-\frac{1}{5})$

= $\boxed{960}$

3600 = 2^5 . 3^2 . 5^2. Thus, any number not greater than 3600 that is not coprime to it is divisible by at least one of 2, 3 or 5.

Using the addition rule for 3 set, where: Set A: numbers divisible by 2 and not greater than 3600, n(A) = 1800 Set B: numbers divisible by 3 and not greater than 3600, n (B) = 1200 Set C: numbers divisible by 5 and not greater than 3600, n(C) = 720

To use the addition rule, we first find A intersect B = multiples of 6 not greater than 3600, n(A intersect B) = 600 A intersect C = multiples of 10 not greater than 3600, n(A intersect C) = 360 B intersect C = multiples of 15 not greater than 3600, n(B intersect C) = 240 A intersect B intersect C = multiples of 30 not greater than 3600, n(A intersect B intersect C) = 120

Thus, numbers coprime to 3600 and not greater than 3600 = n(A U B U C) = n(A) + n(B) + n (C) - n (A intersect B) - n (A intersect C) - n(B intersect C) + n( A intersect B intersect C) = 1800 + 1200 + 720 - 600 - 360 - 240 + 120 = 2640

Hence, the answer we want is 3600 - 2640 = 960