
PreCalculus PreCalculus Math Forum 
 LinkBack  Thread Tools  Display Modes 
February 23rd, 2018, 06:16 PM  #11 
Senior Member Joined: Feb 2010 Posts: 697 Thanks: 135  
February 24th, 2018, 11:02 AM  #12  
Senior Member Joined: May 2016 From: USA Posts: 1,210 Thanks: 498  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, 12:25 PM  #13 
Senior Member Joined: Feb 2010 Posts: 697 Thanks: 135 
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, 12:37 PM  #14  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,641 Thanks: 954  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 12:51 PM.  
February 24th, 2018, 01:03 PM  #15 
Senior Member Joined: May 2016 From: USA Posts: 1,210 Thanks: 498 
You are asking ME about how things get parsed?

February 24th, 2018, 01:12 PM  #16  
Member Joined: Jun 2017 From: Lima, Peru Posts: 52 Thanks: 1 Math Focus: Calculus  Quote:
Quote:
 
February 24th, 2018, 02:05 PM  #17  
Senior Member Joined: May 2016 From: USA Posts: 1,210 Thanks: 498  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.  

Tags 
common, elements, find, involving, multiples, number, problem 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
How to find the sum of the multiples  popoder  Algebra  5  March 6th, 2016 06:20 AM 
Group Theory Proofs, least common multiples, and group operations HELP!  msv  Abstract Algebra  1  February 19th, 2015 12:19 PM 
Find number of elements combinations covered by a given set  grigor  Applied Math  0  March 26th, 2014 12:25 AM 
Common multiples of two numbers  md9  Abstract Algebra  1  May 10th, 2013 02:26 PM 
Common elements  dajaka  Number Theory  3  October 30th, 2008 02:25 PM 