Unique Percentage

Number Theory Level 3

Let x < y x <y be positive integers such that gcd ⁡ ( x , y ) = 1 \gcd(x,y)=1 . We define

f ( x , y ) = ⌊ x y × 100 ⌋ 100 f(x,y) = \dfrac{\left\lfloor \dfrac{x}{y} \times 100 \right\rfloor}{100}

What is the greatest integer G G , such that for any x 1 , x 2 , y 1 , y 2 ≤ G x_1,x_2, y_1,y_2 \leq G satisfying x 1 ≠ x 2 , y 1 ≠ y 2 x_1 \neq x_2, y_1 \neq y_2 , we have

f ( x 1 , y 1 ) ≠ f ( x 2 , y 2 ) ? f(x_1, y_1) \neq f(x_2, y_2 ) ?


The answer is 10.

Michael Huang by Michael Huang , 29, USA

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1 solution

Alex Burgess
Apr 29, 2019

b 2 3 4 5 6 7 8 9 10 11 a ⌊ 100 a b ⌋ 1 50 33 25 20 16 14 12 11 10 09 2 66 40 28 22 18 3 75 60 42 37 30 27 4 80 56 44 36 5 83 71 62 55 45 6 85 54 7 87 77 70 63 8 88 72 9 9 0 ∗ 81 10 9 0 ∗ \begin{matrix} & b & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 \\ & a\\ \left \lfloor{100\frac{a}{b}} \right \rfloor& 1 & 50 & 33 & 25 & 20 & 16 & 14 & 12 & 11 & 10 & 09 \\ & 2 &&66 &&40&&28&&22&&18\\ &3&&&75&60&&42&37&&30&27\\ &4&&&&80&&56&&44&&36\\ &5&&&&&83&71&62&55&&45\\ &6&&&&&&85&&&&54\\ &7&&&&&&&87&77&70&63\\ &8&&&&&&&&88&&72\\ &9&&&&&&&&& 90^* &81\\ &10&&&&&&&&&& 90^* \\ \end{matrix}

⌊ 100 9 10 ⌋ = ⌊ 100 10 11 ⌋ = 90 \left \lfloor{100\frac{9}{10}} \right \rfloor = \left \lfloor{100\frac{10}{11}} \right \rfloor = 90 is the first duplicate.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

If there is a restriction that all x 1 , x 2 , y 1 , y 2 x_1,x_2,y_1,y_2 are unique we must look further: b = 12 b = 12 only offers 100 12 = 08 , 500 12 = 41 , 700 12 = 58 , 1100 12 = 91 \frac{100}{12} = 08, \frac{500}{12} = 41, \frac{700}{12} = 58, \frac{1100}{12} = 91 .

b = 13 b r i n g s 07 , 15 , 23 , 3 0 ∗ , 38 , 46 , 53 , 61 , 69 , 76 , 84 , 92 b = 13 brings 07, 15, 23, 30^*, 38, 46, 53, 61, 69, 76, 84, 92 with

⌊ 100 4 13 ⌋ = ⌊ 100 3 10 ⌋ = 30 \left \lfloor {100 \frac{4}{13}} \right \rfloor = \left \lfloor {100 \frac{3}{10}} \right \rfloor = 30 .

Alex Burgess - 2 years, 1 month ago

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