Determine the number of positive integers such that $[\frac{a}{2}] + [\frac{a}{3}] + [\frac{a}{5}] = a$ Problem:
Determine the number of positive integers such that:
$$\Big[\frac{a}{2}\Big] + \Big[\frac{a}{3}\Big] + \Big[\frac{a}{5}\Big] = a$$
(where [x] is the greatest integer function)
Attempted solution:
The greatest integer function is essentially the floor function. The largest integer that is less than x.
In order for three numbers to sum up to an integer, they can:


*

*all be integers

*one of them are integers and the remaining two add up to become an integer


If a is even, then
$$\Big[\frac{a}{2}\Big]$$
is also even and we could remove the floor function, since the result is an even integer. This is not possible of a is odd. Similar method can be used for the others, assuming they are divisible by 3 and divisible by 5.
I am not entirely sure how to continue. I thought about replacing a with $2n$ or $2n+1$ in the two different scenarios, but I am not sure that will work since it looks like it just reduces to $a = 0$.
There does not seem to be a general solution methodology for floor functions in the same way that there are for the absolute value function?
The number of positive integers turns out to be 30 according to Wolfram Alpha, which is suspiciously identical to $5*3*2 = 30$ This would perhaps indicate that one should put the left-hand side together on the same fraction, but I am not entirely clear how this is done with floor functions.
 A: $$ \frac{a}{2} + \frac{a}{3} + \frac{a}{5}-3<\Big[\frac{a}{2}\Big] + \Big[\frac{a}{3}\Big] + \Big[\frac{a}{5}\Big]=a\leq \frac{a}{2} + \frac{a}{3} + \frac{a}{5}$$
A: Hint: use $$x-1<[x]\leq x$$ then you get $$ \frac{a}{2} + \frac{a}{3} + \frac{a}{5}-3\leq a\leq \frac{a}{2} + \frac{a}{3} + \frac{a}{5}$$
I get 29 solutions...
A: Hint
Let $a=30k+r$ where $k\in \Bbb N\cup\{0\}$ and $0\le r<30$. Therefore$$31k+\Big[\frac{r}{2}\Big] + \Big[\frac{r}{3}\Big] + \Big[\frac{r}{5}\Big] = 30k+r$$or$$k+\Big[\frac{r}{2}\Big] + \Big[\frac{r}{3}\Big] + \Big[\frac{r}{5}\Big] = r$$
A: Sorry,I just made a mistake.
I think the answer is 30.
when a=2k,$k\in N$;
Compared to$a\over2$+$a\over3$+$a\over5$,the left side may less $i\over3$+$j\over5$(i=0,1,2;j=0,1,2,3,4),
so there are 1x3x5-1=14 numbers fitting it.(Because what the left side less will equal to($a\over2$+$a\over3$+$a\over5$-$a$) ,and i,j cannot equals to 0 at the same time as $a\neq0$)
when a=2k+1,$k\in N$;
Compared to$a\over2$+$a\over3$+$a\over5$,the left side may less $1\over2$+$i\over3$+$j\over5$(i=0,1,2;j=0,1,2,3,4),so there are 1x3x5=15 numbers fitting it.
Above all,14+15=29 numbers.
A: Since
$$
\left\lfloor x \right\rfloor  = x - \left\{ x \right\}\quad \left| {\,0 \le \left\{ x \right\} < 1} \right.
$$
then the equation becomes
$$
\eqalign{
  & 0 = \left\lfloor {{a \over 2}} \right\rfloor  + \left\lfloor {{a \over 3}} \right\rfloor  + \left\lfloor {{a \over 5}} \right\rfloor  - a =   \cr 
  &  = {a \over 2} + {a \over 3} + {a \over 5} - a - \left\{ {{a \over 2}} \right\} - \left\{ {{a \over 3}} \right\} - \left\{ {{a \over 5}} \right\} =   \cr 
  &  = {a \over {30}} - \left\{ {{a \over 2}} \right\} - \left\{ {{a \over 3}} \right\} - \left\{ {{a \over 5}} \right\} \cr} 
$$
Actually, for integral $a$,  we have
$$
\left\{ {{a \over 2}} \right\} \le {1 \over 2},\quad \left\{ {{a \over 3}} \right\} \le {2 \over 3},\quad \left\{ {{a \over 5}} \right\} \le {4 \over 5}
$$
so a first result we get
$$
{a \over {30}} = \left\{ {{a \over 2}} \right\} + \left\{ {{a \over 3}} \right\} + \left\{ {{a \over 5}} \right\}\quad  \Rightarrow \quad 0 \le a < 90
$$
Consider values of $a$ which are multiples of the three factors
$$
\eqalign{
  & {{2b} \over {30}} = \left\{ {{{2b} \over 3}} \right\} + \left\{ {{{2b} \over 5}} \right\} \le \left\{ {{2 \over 3}} \right\} + \left\{ {{4 \over 5}} \right\}
 = {{22} \over {15}}\quad  \Rightarrow \quad 0 \le a_{\,mult2}  \le 44  \cr 
  & {{3c} \over {30}} = \left\{ {{{3c} \over 2}} \right\} + \left\{ {{{3c} \over 5}} \right\} \le \left\{ {{1 \over 2}} \right\} + \left\{ {{3 \over 5}} \right\}
 = {{11} \over {10}}\quad  \Rightarrow \quad 0 \le a_{\,mult3}  \le 33  \cr 
  & {{5d} \over {30}} = \left\{ {{{5d} \over 2}} \right\} + \left\{ {{{5d} \over 3}} \right\} \le \left\{ {{1 \over 2}} \right\} + \left\{ {{2 \over 3}} \right\}
 = {5 \over 6}\quad  \Rightarrow \quad 0 \le a_{\,mult5}  \le 25 \cr} 
$$
which means that you can do some sieving in testing the values of $a$ from $0$ to $59$.
