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Find-the-number-of-rational-numbers-r-0-lt-r-lt-1-such-that-when-r-is-written-as-a-fraction-in-lowest-term-The-numerator-and-the-denominator-have-a-sum-of-1000-




Question Number 111155 by Aina Samuel Temidayo last updated on 02/Sep/20
Find the number of rational numbers  r, 0<r<1, such that when r is written  as a fraction in lowest term. The  numerator and the denominator  have a sum of 1000.
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{rational}\:\mathrm{numbers} \\ $$$$\mathrm{r},\:\mathrm{0}<\mathrm{r}<\mathrm{1},\:\mathrm{such}\:\mathrm{that}\:\mathrm{when}\:\mathrm{r}\:\mathrm{is}\:\mathrm{written} \\ $$$$\mathrm{as}\:\mathrm{a}\:\mathrm{fraction}\:\mathrm{in}\:\mathrm{lowest}\:\mathrm{term}.\:\mathrm{The} \\ $$$$\mathrm{numerator}\:\mathrm{and}\:\mathrm{the}\:\mathrm{denominator} \\ $$$$\mathrm{have}\:\mathrm{a}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{1000}. \\ $$
Answered by Her_Majesty last updated on 02/Sep/20
r=(n/(1000−n)); 0<r<1 ⇒ 0<n<500∧n≠2k∧n≠5k  between 0 and 500: 249 numbers n=2k and  99 numbers n=5k but we have 49 numbers  n=2×5k ⇒ 50 numbers left  sum of not allowed numbers = 299  ⇒ allowed numbers =200 =answer
$${r}=\frac{{n}}{\mathrm{1000}−{n}};\:\mathrm{0}<{r}<\mathrm{1}\:\Rightarrow\:\mathrm{0}<{n}<\mathrm{500}\wedge{n}\neq\mathrm{2}{k}\wedge{n}\neq\mathrm{5}{k} \\ $$$${between}\:\mathrm{0}\:{and}\:\mathrm{500}:\:\mathrm{249}\:{numbers}\:{n}=\mathrm{2}{k}\:{and} \\ $$$$\mathrm{99}\:{numbers}\:{n}=\mathrm{5}{k}\:{but}\:{we}\:{have}\:\mathrm{49}\:{numbers} \\ $$$${n}=\mathrm{2}×\mathrm{5}{k}\:\Rightarrow\:\mathrm{50}\:{numbers}\:{left} \\ $$$${sum}\:{of}\:{not}\:{allowed}\:{numbers}\:=\:\mathrm{299} \\ $$$$\Rightarrow\:{allowed}\:{numbers}\:=\mathrm{200}\:={answer} \\ $$$$ \\ $$
Commented by Aina Samuel Temidayo last updated on 02/Sep/20
Thanks
$$\mathrm{Thanks} \\ $$

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