If $a(a-\{3a\})$ is an integer, then $a(\lfloor 3a \rfloor -2a)$ is an integer. For $a \in (3,\frac{10}{3})$, then $a(9-2a) \in \mathbb{Z}$. No solution. For $a \in [\frac{10}{3},\frac{11}{3})$ $$2a(5-a) \in \mathbb{Z}$$COnfused....

$a=3+\{a\}$ For $3<a<\frac{10}{3}: a(a-\{3a\})=a([3a]-2a)=a(9-2a)< \frac{100}{9} \to a(9-2a)=9,10,11$ - there are not solutions. For $a \in [\frac{10}{3},\frac{11}{3}): a([3a]-2a)=a(10-2a)=10,11 \to a=\frac{5+\sqrt{5}}{2},\frac{5+\sqrt{3}}{2}$ For $a \in [\frac{11}{3},4): a([3a]-2a)=a(11-2a)=13 \to a=\frac{11+\sqrt{17}}{4}$

parmenides51 wrote: Find all real numbers $a$ such that $3 < a < 4$ and $a(a-\{3a\})$ is an integer. (Here $\{3\}$ denotes the fractional part of $a$.) I think that $\{3\}$ should be $\{a\}$

parmenides51 wrote: Find all real numbers $a$ such that $3 < a < 4$ and $a(a-\{3a\})$ is an integer. (Here $\{3\}$ denotes the fractional part of $a$.) I think it should $a(a-3\{a\})\in\mathbb{Z}$ according to official CRMO paper

Firstly, note that the question posted here is wrong. In the original question, it was $a(a-3\{a\})$ and not $a(a-\{3a\})$. So, proceeding with the corrected question, it's a simple sum.. $\big(\because a\in(3,4)\big)$, consider $a=3+x$, where $x\in(0,1)$. $\therefore a(a-3\{a\})\in\Bbb{Z}$ $\implies (3+x)(3-2x)\in\Bbb{Z}$ $\implies 9-(2x^2+3x)\in\Bbb{Z}$ $\implies (2x^2+3x)\in\Bbb{Z}$ Now, we know $x\in(0,1)$ $\implies (2x^2+3x)\in(0,5)$. That is, it can take values $\{1,2,3,4\}$ Solving the equations - $2x^2+3x=1$ $2x^2+3x=2$ $2x^2+3x=3$ $2x^2+3x=4$ We get the feasible solutions as - $x= \frac{\sqrt{33}-3}{4},$ $\frac{1}{2},$ $\frac{\sqrt{41}-3}{4},$ $\frac{\sqrt{17}-3}{4}$ So, $a= 3+\frac{\sqrt{33}-3}{4},$ $\frac{7}{2},$ $3+\frac{\sqrt{41}-3}{4},$ $3+\frac{\sqrt{17}-3}{4}$ This completes the answer..

parmenides51 wrote: Find all real numbers $a$ such that $3 < a < 4$ and $a(a-\{3a\})$ is an integer. (Here $\{3\}$ denotes the fractional part of $a$.) Math-wiz wrote: I think that $\{3\}$ should be $\{a\}$ Jupiter_is_BIG wrote: I think it should $a(a-3\{a\})\in\mathbb{Z}$ according to official CRMO paper stranger_02 wrote: Firstly, note that the question posted here is wrong. In the original question, it was $a(a-3\{a\})$ and not $a(a-\{3a\})$. these 2 typos has just been corrected

parmenides51 wrote: parmenides51 wrote: Find all real numbers $a$ such that $3 < a < 4$ and $a(a-\{3a\})$ is an integer. (Here $\{3\}$ denotes the fractional part of $a$.) Math-wiz wrote: I think that $\{3\}$ should be $\{a\}$ Jupiter_is_BIG wrote: I think it should $a(a-3\{a\})\in\mathbb{Z}$ according to official CRMO paper stranger_02 wrote: Firstly, note that the question posted here is wrong. In the original question, it was $a(a-3\{a\})$ and not $a(a-\{3a\})$. these 2 typos has just been corrected Good..

We can write $\{a\}=a-[a]$ where $[a]$ is the greatest integer less than or equal to $a$. $a(a-3(a-[a])\\ = -2a^2+3a[a]\\ =9a-2a^2 \in \mathbb{Z}$ We can take $[a]=3$ as $3<a<4$. This means: $2a^2-9a=-k \in \mathbb{Z}\\ \iff 2a^2-9a+k$ has a real solution and the solution is between $3$ and $4$. $3 < \frac{9\pm \sqrt{81-8k}}{4} < 4\\ \iff 4<k<9$ Hence all the values of $a$ are roots of $2a^2-9a+k$ where $k \in {5,6,7,8}$