Math Book Cambodia: Prove that: `1/2+1/5+1/8+1/11+1/20+1/40+1/110+1/1640=1`

Sunday, March 7, 2021

1. Prove that: $1/2+1/5+1/8+1/11+1/20+1/40+1/110+1/1640=1$

Solution

We observe that:

$1/2+1/5+1/8+1/11+1/20+1/40+1/110+1/1640$

$=1/2+1/20+1/5+1/8+1/11+1/110+1/20+1/1640$

$=22/40+13/40+11/110+41/1640$

$=22/40+13/40+4/40+1/40$

$=40/40=1$

Hence, $1/2+1/5+1/8+1/11+1/20+1/40+1/110+1/1640=1$

2. Find the minimum value of $A=|x-1|+|x-2|+|x-3|+.....+|x-100|$

Solution

We observe that:

$|x-k|+|x-(101-k)|$ $\geq$ $|101-2k|$ while $|a|+|b|$ $\geq$ $|a-b|$

The equal case happens when $x\in$ $[x,101-k]$

For all $k=1,2,3,....,50$ we will get that:

$|x-1|+|x-100|$ $\geq$ $99$

$|x-2|+|x-99|$ $\geq$ $97$

$|x-3|+|x-98|$ $\geq$ $95$

$.................................................................$

$|x-50|+|x-51|$ $geq$ $1$

Make the sum of side to side, we will get:

$A\geq$ $99+97+95+......+1$ $=50/2(1+99)$ $=2500$

The equal case happens when $x\in$ $\cap[k,101-k]$ which $1$ $\geq$ $k$ $\geq$ $501$ $=[50,51]$

Hence, the minimum value of $A=|x-1|+|x-2|+|x-3|+.....+|x-100|=2500$ when $x\in$ $=[50,51]$

$A_{min}=2500$ when $x\in$ $=[50,51]$

Math Book Cambodia