Problem of the week #245 Dec 14th, 2016

[anemone]

Here is this week's POTW:
-----
Prove that $$\frac{aca}{acb}\lt \frac{bca}{bcb}$$ for any digits $a\ne b$ and for any digit number $c$, where $xyz$ represents a 3-digit number.
-----
Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to the following members for their correct solution::)

1. kaliprasad
2. IanCg

Solution from IanCg:

Spoiler

From the given inequality we have

$\dfrac{100a+10c+a}{100a+10c+b}<\dfrac{100b+10c+a}{100b+10c+b}$

It's true if $(100a + 10c + a)(100b + 10c +b) < (100b + 10c + a)(100a + 10c + b)$.

Expanding both sides we get:

$10000ab+1000ac+100ab+1000bc+100{c}^{2}+10bc +100ab+10ac+ab<10000ab+1000bc+100{b}^{2}+1000ac+100{c}^{2}+10bc+100{a}^{2}+10ac+ab$

Cancelling like terms from each side gives

$200ab<100{b}^{2}+100{a}^{2}$
$2ab<{b}^{2}+{a}^{2}$
$0<(a-b)^{2}$, which is true since the question said $a$ was not equal to $b$.