I was solving questions from a book and it had a question :
Find all two digit numbers such that the sum of digits constituting the number is not less than 7; the sum of squares of digits is not greater than 30; the number written in reverse order is not larger than half the numbers ".
I proceeded like this:
Let the no be $10x+y$.
So from the questions we have :
$$x+y \ge 7 \tag{1}$$
$$x^2 + y^2 \le 30\tag{2}$$
$$10y+x \le \frac{1}{2}(10x+y)\tag{3}$$
where $x \in [1,9]$ and $y\in[0,9]$. Now I am stuck up. How to proceed now?
Best Answer
$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $$ 7^{2} \leq x^{2} + 2xy + y^{2}\leq 30 + 2xy\quad\imp\quad -2xy\leq -19 $$
$$-3\leq x - y\leq 3$$
Also, $\quad\ds{x \geq {19 \over 8}\,y\geq {19 \over 4}\quad\imp\quad \color{#66f}{\Large x\geq 5}.\quad}$
Also, $\ds{30 \geq x^{2} + y^{2}\geq 25 + y^{2}\ \imp\ \color{#66f}{\Large \verts{y} \leq \root{5}}}$ .Then $\color{#66f}{\Large \ds{y = 2}}$.
There is just one solution: $\ds{\color{#c00000}{\LARGE 52}}$.