MHB Problem of the week #245 Dec 14th, 2016

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The Problem of the Week #245 challenges participants to prove that \(\frac{aca}{acb} < \frac{bca}{bcb}\) for any distinct digits \(a\) and \(b\), and any digit number \(c\). The discussion highlights the importance of understanding the properties of three-digit numbers represented as \(xyz\). Participants are encouraged to submit their solutions following the established guidelines. Notable contributors who provided correct solutions include kaliprasad and IanCg. The thread emphasizes engagement with the mathematical community through problem-solving.
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Here is this week's POTW:
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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.
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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!
 
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Congratulations to the following members for their correct solution::)

1. kaliprasad
2. IanCg

Solution from IanCg:

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 $b$.
 
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