Who Has What? Solving Inequalities in Number Theory

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Anna, Betty, and Carl collectively have $100, leading to the equation A + B + C = 100. Betty introduces a condition where if Anna had six times her amount and she had one-third of her amount, the total would still be $100, resulting in the equation 6A + (1/3)B + C = 100. By manipulating these equations, it is determined that B equals (15/2)A, implying that Anna must have an even dollar amount. Additionally, Carl's amount is constrained to be less than $30, leading to the inequality 0 < 200 - 17A < 60. The problem requires finding integer solutions that satisfy these conditions.
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Anna says, "We three have $100 altogether." Betty says, "Yes, and if you had six times as much and I had one-third as much, we three would still have $100." Carl says, It's not fair. I have less than $30." Who has what? (Dudley)
 
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This being a textbook problem, the question belongs in the Homework & Coursework forum.
 
Let A, B, C be the amounts of money Anna, Betty, and Carl have, respectively.

"We three have $100 altogether." A+ B+ C= 100

Betty (presumably responding to Anna) says, "if you had six times as much and I had one-third as much, we three would still have $100"
6 times as much for Anna would be 6A and one-third as much for Betty would be (1/3)B: 6A+ (1/3)B+ C= 100. Subtracting those two equations will eliminate C (Carl): (6-1)A+ (1/3- 1)B= 5A- 2/3 B= 0 or 2/3B= 5A so B= (15/2)A. Now, I think we will have to assume that each person has an integer number of dollars- in particular it follows from that that Anna has an even number of dollars. Use that equation, together with the fact that Carl has less than 30 dollars to see what integer amounts will satisfy the conditions.
 
Note that 0\leq{C}=100-\frac{2}{2}A-\frac{15}{2}A=\frac{200-17A}{2}&lt;30
Thus, you have the inequality:
0<200-17A<60
 
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