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February 23rd, 2018, 05:16 PM  #11 
Senior Member Joined: Feb 2010 Posts: 683 Thanks: 129  
February 24th, 2018, 10:02 AM  #12  
Senior Member Joined: May 2016 From: USA Posts: 1,084 Thanks: 446  Quote:
$n =89.$ $a = 29 \implies 3 * 29 = 87 = 89  2.$ $b = 12 \implies 7 * 12 = 84 = 89  5.$ $c = 10 \implies 8 * 10 = 80 = 89  9.$ Here is how I did it. Problem find the smallest possible n subject to the following conditions: $a,\ b,\ c,\ n \in \mathbb N+$ $n = 3a + 2.$ $n = 7b + 5.$ $n = 8c + 9.$ Solution. We have three equations and four unknowns, but have extra information in that we are looking for the smallest integer solution for n. It is obvious that to minimize n, we must minimize c. $3a + 2 = n = 8c + 9 \implies 3a = 8c + 7 = 9c + 9  (c + 2) \implies a = 3c + 3  \dfrac{c + 2}{3}.$ For a to be an integer, c + 2 must be divisible by 3. In other words, we need an integer p such that c = 3p  2. $7b + 5 = n = 8c + 9 \implies b = \dfrac{8c + 4}{7} = c + \dfrac{c + 4}{7}.$ For b to be an integer, c + 4 must be divisible by 7. In other words, we need an integer q such that c = 7q  4. And we want the smallest such values of p and q that give an integer value to c. $7q  4 = 3p  2 \implies 7q = 3p + 2 \implies q = \dfrac{3p + 2}{7}.$ We now have 1 equation in two unknowns. There may be fancy ways to solve it in positive integers, but enumeration works. $p = 1 \implies \dfrac{3p + 2}{7} = \dfrac{5}{7} \implies q \not \in \mathbb Z.$ $p = 2 \implies \dfrac{3p + 2}{7} = \dfrac{8}{7} \implies q \not \in \mathbb Z.$ $p = 3 \implies \dfrac{3p + 2}{7} = \dfrac{11}{7} \implies q \not \in \mathbb Z.$ $p = 4 \implies \dfrac{3p + 2}{7} = \dfrac{14}{7} \implies q \in \mathbb Z.$ Now here we go. $c = 3p  2 = 10 \implies$ $n = 8 * 10 + 9 = 89 \implies$ $7b + 5 = 89 \implies 7b = 84 \implies b = 12.$ $3a + 2 = 89 \implies 3a = 87 \implies a = 29.$  
February 24th, 2018, 11:25 AM  #13 
Senior Member Joined: Feb 2010 Posts: 683 Thanks: 129 
Jeff ... I'm no expert in number theory! I just followed the algorithm in an old book I have. The algorithm gave an answer of 929 mod 168. This reduces to 89 mod 168. So in fact all possible answers are $\displaystyle 89+168t$ for any integer value of $\displaystyle t$. Some day when my brain is working, I'll try to read the background on this to see why it works. I actually tried to do it the way you did but I got tangled up so I fell back on the algorithm.

February 24th, 2018, 11:37 AM  #14  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,904 Thanks: 883  Quote:
$a = 29 \implies 3 * 29 = 87 = 89  2.$ Jeff, I keep getting these "yellow weirdoes" when you guys use the $ in order to format: not just from you, others also... Is there something I need to change from my end? Last edited by Denis; February 24th, 2018 at 11:51 AM.  
February 24th, 2018, 12:03 PM  #15 
Senior Member Joined: May 2016 From: USA Posts: 1,084 Thanks: 446 
You are asking ME about how things get parsed?

February 24th, 2018, 12:12 PM  #16  
Member Joined: Jun 2017 From: Lima, Peru Posts: 46 Thanks: 1 Math Focus: Calculus  Quote:
Quote:
 
February 24th, 2018, 01:05 PM  #17  
Senior Member Joined: May 2016 From: USA Posts: 1,084 Thanks: 446  Quote:
Once I have reduced the problem to one equation with two integer unknowns, I admit that my method s just guess and check. But because we are looking for a minimum integer, the chances are that finding it will be relatively efficient.  

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