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Total possible = 10 * 10 = 100

pairs when x and y have common factor greater than 1, will be duplicates.
(2,4) (2,6) (2,8) (2,10) (3,6) (3,9) (4,6) (4,8) (4,10) (5,10) (6,8) (6,9) (6,10) (8,10) = 14
Same way when 14 more inverse of these. So total = 14 *2 = 28

9 more duplicates= (2,2)......(10,10) = 9

Answer = 100-28-9 = 63
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Assuming all fractions are unique, the number of fractions in all is 10x10=100

Next, all fractions where the numerator and denominator share a common factor (not 1) are duplicates....

Eliminate
2,2
2,4
2,6
2,8
2,10

(For pairs where x!=y, count it twice, since the inverse is also a duplicate)

totally 9

3,3
3,6
3,9
totally 5

4,4
4,6
4,8
4,10
totally 7

5,5
5,10
totally 3

6,6
6,8
6,9
6,10
totally 7

7,7
totally 1

8,8
8,10
totally 3

9,9
totally 1

10,10
totally 1

total =37

answer = 100 -37 = 63


i probably would miss this one the real exam :(

Anyway, the theory behind this problem is that a uinique fraction x/y must have the property that x,y are different prime numbers or powers of different prime numbers.



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