How many positive values of $a$ are possible in the following case? $$2^{3}\le a\lfloor a\rfloor \le 4^{2} + 1$$ where $a\lfloor a\rfloor$ such that $a[a]$ is an integer.
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It is clear that $\lfloor a\rfloor$ must be $\ge 3$ and $\le 4$. In the case $\lfloor a\rfloor=3$, the product can be $9$, $10$, or $11$. For $8$ is too small, since $a\ge \lfloor a\rfloor$. And $12$ is too big, since then $a=4$, giving the wrong value for the floor function. So $a$ can have values $3$, $10/3$, and $11/3$. In the case $\lfloor a\rfloor=4$ there are two possible values of the product, $16$ and $17$. So the number of possible values of $a$ is $5$. |
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$[a] ≤ a => [a][a] ≤ a[a]≤17 =>[a]≤4$ and $a ≥ [a] => a.a ≥a[a] ≥ 8 =>a≥\sqrt8$ $=>3≤[a]≤4$ If $ [a]=3, 3≤a<4 =>3.3≤a[a]<4.3 => 9≤a[a]<12$ which satisfies the given condition. So, $9≤3a<12=>\frac{9}{3}≤a<\frac{12}{3}=>a=\frac{9}{3}, \frac{10}{3}, \frac{11}{3}$ If $[a]=4, 4≤a<5 =>16≤a[a]<20$, but according to the given condition $8≤a[a]≤17$ So, here $16≤a[a]≤17=>16≤4a≤17=>a=\frac{16}{4}, \frac{17}{4}$ |
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