Two positive real numbers, $a$ and $b$, are expressed as the sum of $m$ positive real numbers and $n$ positive real numbers respectively as follows:
$a = s_1 + s_2 + s_3 + s_4 + \cdots + s_m$
$b = t_1 + t_2 + t_3 + t_4 + \cdots + t_n$
For any real number $x$, $\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x$ and $\lceil x\rceil$ denotes the least integer greater than or equal to $x$.
If $\lfloor a\rfloor = \lceil s_1\rceil + \lceil s_2\rceil + \lceil s_3\rceil + \cdots + \lceil s_m\rceil – 11$ and $\lceil b\rceil = \lfloor t_1\rfloor + \lfloor t_2\rfloor + \lfloor t_3\rfloor + \cdots + \lfloor t_n\rfloor +7$
What is the minimum possible value of $m+n$ ?
My attempt so far :-
I have no clue on what properties to apply , so I started by checking with some values
Since $a>0$ and also $\lfloor a\rfloor \ge 0$
I took a case for minimum value of when $\lfloor a\rfloor = 0$ which would imply $0\le a<1$, now intuitively by taking $s_1, s_2, s_3,\ldots,s_{12}$ to be very small values between $0$ and $1$, I should be able to create the sum to lie between $0$ and $1$, and also each of $12 \lceil s\rceil$ would give value as $1$ which would satisfy our condition, so $m=12$ would be the smallest value
I have no clue on how to fit values for the second set of equations involving $b$ and $t$.
All I can say is $\lceil b\rceil \ge 7$
Also, I am extremely skeptical about my initial approach for finding $\min(m)$ as well.
Please guide me on how to solve this.
Best Answer
Your idea of $m=12$ was reasonable, although you still needed to prove that $m\not<12$! I'm pretty sure my answer is right, but there most likely will be something wrong.
Lets start with $a$! Since $0\leq{a-\lfloor{a}\rfloor}<1$, the expression can be transformed into this: $$a=\lceil{s_1}\rceil+\lceil{s_2}\rceil+\lceil{s_3}\rceil+\cdots+\lceil{s_m}\rceil-k,\ \ \ \ \ \ \ 10<k\leq11$$ Then we can express $\lceil{s_i}\rceil$ as $s_i+\varepsilon_i$, where $0\leq\varepsilon_i<1$. Plugging that in we get: $$a=s_1+\varepsilon_1+s_2+\varepsilon_2+\cdots+s_m+\varepsilon_m-k,\ \ \ \ \ \ \ 10<k\leq11$$ Since $\varepsilon_1+\varepsilon_2+\cdots+\varepsilon_m$ must perfectly cancel out with $-k$, the minimum $m$ value will be $11$.
The procedure for $b$ will be very similar!
Since $0\leq{\lceil{b}\rceil-b}<1$, the expression can be transformed into this: $$b=\lfloor{t_1}\rfloor+\lfloor{t_2}\rfloor+\lfloor{t_3}\rfloor+\cdots+\lfloor{t_n}\rfloor+p,\ \ \ \ \ \ \ 6<p\leq7$$ Then we can express $\lfloor{t_i}\rfloor$ as $t_i-\delta_i$, where $0\leq\delta_i<1$. Plugging that in we get: $$b=t_1-\delta_1+t_2-\delta_2+\cdots+t_n-\delta_n+p,\ \ \ \ \ \ \ 6<p\leq7$$ Since $-\delta_1-\delta_2-\cdots-\delta_n$ must perfectly cancel out with $p$, the minimum $n$ value will be $7$.
Thus we can conclude that the minimum possible value of $m+n$ is $11+7=18$.
Here are two examples for $a$ and $b$!
($a$ 1.) $21.1=1.05+10\cdot2.005$
($a$ 2.) $21=2+10\cdot3-11$
($b$ 1.) $13.9=7\cdot\frac{139}{70}$
($b$ 2.) $14=7\cdot1+7$