Numerator sixty

In the set of possible denominators $\{61, 62, \dots, 9999\}$ we must find those that are coprime to 60.

Since $\gcd(60, a+60b) = \gcd(60, a)$ , the pattern of valid denominators repeats with period sixty. The number of values between $0$ and $59$ that are coprime to 60 is $\phi(60) = 16$ ( Euler totient function ).

Between 60 and 10,000 there are $\lfloor(10\:000 - 60)/60\rfloor = 165$ such periods, accounting $165 \times 16 = 2640$ valid denominators. We only need to look at the denominators beyond the last full period, i.e. $d > 9960$ . Because of symmetry, the first half period ( $9960 \leq d < 9990$ ) contains $16/2 = 8$ valid denominators. Finally, we consider the number of valid denominators between $9990$ and $9999$ , which is the same as the number of values between $30$ and $39$ that are coprime to 60. Only $31$ and $37$ qualify; therefore we have a total of $2640 + 8 + 2 = \boxed{2650}$ denominators that are coprime to 60, and therefore 2650 reduced fractions of the kind required.

2, 3, 4, and 5 are factors of 60. Since what ever divisible by 4 is also divisible by 2, we only consider 2, 3, and 5.
So number with units ending in even digit or 5 or 0 has to be rejected.
From 1 to 30, .........if the tenth digit is 0 unit digit 3 and 9 are rejected ,.................. This is true for tenth digit ç, if X=0. ........Two useful.
...............................if the tenth digit is 1 unit digit 1, 3, 7, and 9 are rejected ,........ This is true for tenth digit T equivalent X (mod 3), if X=1. ........Four useful.
...............................if the tenth digit is 2 unit digit 3 and 9 are rejected ,.................. This is true for tenth digit T equivalent X (mod 3), if X=2. ........Two useful.
So every such group of 30 has 8 useful numbers.
10000/30=333 + 1/3.
333 has 8 useful numbers. This 1/3 is T equivalent 0 (mod 3) , so has Two useful numbers.
Thus 333 * 8 + 2=2666 useful numbers. But 1 to 60 in this are not > 60. So we take away 2 * 8 =16. Reduced positive fractions less than 1, =2666-16= $\Large \color{#D61F06}{2650}.$

We mean to find all integers greater than $60$ and less than $10000$ that are coprime to $60$ , i.e. that are not multiples of any of the divisors of $60$ . Since $60 = 2^{2} * 3 * 5$ , we are looking for all integers in this interval that are not multiples of $2, 3$ or $5$ . One way to do this is to get the number of odd integers in this interval, take out the odd multiples of $5$ , take out the odd multiples of $3$ and finally add the odd multiples of both $3$ and $5$ (multiples of $15$ ), given that we've accounted for them twice in the last two steps. Using the formula $u_{n} = u_{1} + r.(n-1)$ , where n is the number of terms of the AP and r the common ratio, replacing the last term in the interval ( $u_{n}$ ) and the first one yields the number of terms n. STEP 1: ODD NUMBERS: $9999 = 61 + 2.(n-1)$ $\equiv n = 4970$ . STEP 2: MULTIPLES OF 5: $9995 = 65 + 10.(n-1)$ $\equiv n = 994$ . STEP 3: ODD MULTIPLES OF 3: $9999 = 63 + 6.(n-1)$ $\equiv n = 1657$ . STEP 4: ODD MULTIPLES OF 15: $9975 = 75 + 30.(n-1)$ $\equiv n = 331$ . SO, the answer is $4970 - 994 - 1657 + 331 = 2650$ .