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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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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.
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: $\sum_{k=1}^{9}$$99-10k$$=99\cdot9-10\cdot\frac{9}{2}$$9+1$$=891-50\cdot9=441$ I used the wording of the original problem statement where the coefficients are positive, rather than non-negative.

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