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 June 18th, 2011, 08:43 PM #1 Newbie   Joined: Jun 2011 Posts: 5 Thanks: 0 Number of all non-negative solutions. This problem appeared in the British Mathematical Olympiad in 1997. Find total number of positive integral solutions for this equation. a+10b+100c+1000d=1997 The question asks for all combinations of values of a,b,c and d. I'd appreciate help on this. Thanks in advance. June 20th, 2011, 12:52 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,931 Thanks: 2205 Clearly the last digit of a is 7, b < 90, c < 10, and d = 1. Have you tried to find the total when, for example, c = 9? Do that, then try c = 8, etc. June 20th, 2011, 02:53 AM   #3
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 Originally Posted by skipjack Clearly the last digit of a is 7, b < 90, c < 10, and d = 1. Have you tried to find the total when, for example, c = 9? Do that, then try c = 8, etc.
Isn't a = 97, b = 80, c = 1, d = 1 a solution? June 20th, 2011, 02:59 AM   #4
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Originally Posted by mrtwhs
Quote:
 Originally Posted by skipjack Clearly the last digit of a is 7, b < 90, c < 10, and d = 1. Have you tried to find the total when, for example, c = 9? Do that, then try c = 8, etc.
Sorry! You are correct. I misread b < 90 to say a < 90  June 22nd, 2011, 06:04 AM #5 Newbie   Joined: Jun 2011 Posts: 5 Thanks: 0 Re: Number of all non-negative solutions. The problem is not finding individual solutions of the equation. The problem lies in finding the NUMBER OF ALL POSSIBLE SOLUTIONS. If there were no coefficients i.e a+b+c+d=1997 the total number of all integral nonnegative solutions can be found through (1997+4-1)C(1997). But the coefficients are a stumbling block for me. June 22nd, 2011, 06:42 AM #6 Global Moderator   Joined: Dec 2006 Posts: 20,931 Thanks: 2205 Can you find the few individual solutions when c = 9 and d = 1? July 31st, 2011, 06:28 AM #7 Newbie   Joined: Jun 2011 Posts: 5 Thanks: 0 Re: Number of all non-negative solutions. Its a diophantine equation with 4 variables. Find the general formula for each variable and then restrict them to nonnegative values. July 31st, 2011, 07:06 AM #8 Senior Member   Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Number of all non-negative solutions. With d = 1, the problem reduces to: a+10b+100c = 997 For each c, where 1 ? c < 10, there are 99 - 10c solutions, thus the total number of solutions S is: I used the wording of the original problem statement where the coefficients are positive, rather than non-negative. Tags nonnegative, number, solutions Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post scarecr0w132 Elementary Math 3 February 24th, 2014 08:19 AM bidelco Number Theory 3 February 14th, 2013 08:44 AM wonderwall Number Theory 7 April 23rd, 2012 12:42 PM sharp Algebra 5 December 23rd, 2010 09:00 PM zain Elementary Math 2 April 26th, 2008 01:07 PM

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