parmenides51 wrote: For all positive integer $n$, define $f_n(x)$ such that $f_n(x) = \sum_{k=1}^n{|x - k|}$. Determine all solution from the inequality $f_n(x) < 41$ for all positive $2$-digit integers $n$ (in decimal notation). Easy to see that $f_n(x)$ is a continuous sequence of segments whose slopes are $-n,2-n,4-n,...,n-4,n-2,n$ Note also that $f_n(x)=f_n(n+1-x)$ and so $f_n(x)$ graph is symetric versus $x=\frac{n+1}2$ 1) If $n=2m+1$ is odd (with $m\in\{5,6,7,..., 49\}$ in order $n$ have two digits) Minimum is reached when slope moves from $-1$ (on $(m,m+1$ ) to $+1$ (on $(m+1,m+2)$ ) So minimum is $f_{2m+1}(m+1)=\sum_{k=1}^{2m+1}|m+1-k|$ $=\sum_{k=1}^m(m+1-k)+\sum_{k=m+2}^{2m+1}(k-m-1)$ $=m(m+1)$ If $m\ge 6$ this means $f_{2m+1}(x)\ge 42$ $\forall x$ and no solution So we need $m=5$ and it is easy to check that : $f_{11}(x)=x+24$ over $[6,7]$ with values in $[30,31]$ $f_{11}(x)=3x+10$ over $[7,8]$ with values in $[31,34]$ $f_{11}(x)=5x-6$ over $[8,9]$ with values in $[34,39]$ $f_{11}(x)=7x-24$ over $[9,10]$ with values in $[39,46]$ (slopes increase $+2$ at each step and constants are computed just with continuity) And so $x<\frac{65}7$ And (symetry versus $x=6$), $x>\frac{19}7$ And solution $\boxed{\text{S1 : }n=11\text{ and }x\in\left(\frac{19}7,\frac{65}7\right)}$ 2) If $n=2m$ is even (with $m\in\{5,6,7,..., 49\}$ in order $n$ have two digits) Minimum is reached when slope reaches $0$ (on $(m,m+1)$ ) So minimum is $f_{2m}(m)=f_{2m}(m+1)=\sum_{k=1}^{2m}|m-k|=m^2$ if $m\ge 7$, this means $f_{2m}(x)\ge 49$ and no solution. So $m\in\{5,6\}$ 2.1) If $m=5$ and so $n=10$ $f_{10}(x)=25$ over $(5,6)$ $f_{10}(x)=2x+13$ over $(6,7)$ with values in $(25,27)$ $f_{10}(x)=4x-1$ over $(7,8)$ with values in $(27,31)$ $f_{10}(x)=6x-17$ over $(8,9)$ with values in $(31,37)$ $f_{10}(x)=8x-35$ over $(9,10)$ with values in $(37,45)$ (slopes increase $+2$ at each step and constants are computed just with continuity) And so $x<\frac{19}2$ And (symetry versus $x=\frac{11}2$), $x>\frac 32$ And solution $\boxed{\text{S2 : }n=10\text{ and }x\in\left(\frac 32,\frac{19}2\right)}$ 2.2) If $m=6$ and so $n=12$ $f_{12}(x)=36$ over $(6,7)$ $f_{12}(x)=2x+22$ over $(7,8)$ with values in $(36,38)$ $f_{12}(x)=4x+6$ over $(8,9)$ with values in $(38,42)$ (slopes increase $+2$ at each step and constants are computed just with continuity) And so $x<\frac{35}4$ And (symetry versus $x=\frac{13}2$), $x>\frac {17}4$ And solution $\boxed{\text{S3 : }n=12\text{ and }x\in\left(\frac {17}4,\frac{35}4\right)}$