Math Goodies is a free math help portal for students, teachers, and parents.
|
Interactive Math Goodies Software

testing left nav
Math Forums @ Math Goodies
Home | Profile | Register | Active Topics | Members | Search | FAQ
 All Forums  Homework Help Forums  Miscellaneous Math Topics  Stamp Collector New Topic  Reply to Topic  Printer Friendly
Author  Topic

TchrWill

USA
79 Posts

 Posted - 03/24/2013 :  08:06:48 Joe came back from a stamp collectors gathering and told his sister, Jill, that he bought a hundred stamps. Joe said he bought the stamps at four different prices: \$0.59, \$1.99, \$2.87, and \$3.44 each. Jill asked, "How much did you pay altogether?" Joe replied, "One hundred dollars exactly." How many stamps did Joe buy at each price? EnjoyI am sure that others would appreciate it if you would post your solution paths so that we might all benefit from seeing varying techniques that are used in solving these types of problems.Many thanks for your participation.Bill

someguy

143 Posts

 Posted - 03/24/2013 :  17:48:50 Hi TchrWill.Let A be the number of 59 cent stamps.Let B be the number of 199 cent stamps.Let C be the number of 287 cent stamps.Let D be the number of 344 cent stamps.We are told that A+B+C+D = 10059A+199B+287C+344D=10000Substituting A=100-B-C-D into the second equation gives140B + 228C + 285D = 4100Note that 59 divides both 228 and 285.Since 59 = 3*19, we can mod out by 3, then by 19, and then combine the results.2B is congruent to -1 mod 37B is congruent to -4 mod 19This meansB = 1 (mod 3)B = 13 (mod 19)Putting these together leads to B=13 (mod 59).Since 1.99B cant be negative and cant be more than 100, we see that B must be 13.This leads toB=13228C + 285D = 2280We get lucky. Note that C=10, D=0 solves this equation.Since A=100-B-C-D, A=77.He bought 77 59 cent stamps, 13 199 cent stamps, 10 287 cent stamps, and 0 344 cent stamps.As a quick check,77+13+10+0=10077(.59) + 13(1.99) + 10(2.87) + 0(3.44) = 100.00Note: if we had not 'spotted the solution', we could have used the last equationinvolving only C and D to say D is a multiple of 4.Since we need A to be at least 50 and B is 13, the fact that the price is \$100lets us know D cant be greater than 12. If we had not spotted the solution,we would have said D=0, 4, 8, or 12 and just checked these.

TchrWill

USA
79 Posts

 Posted - 03/24/2013 :  22:23:25 Nice touch someguy, if no \$3.44 value stamps were bought. The problem statement asks "How many stamps did Joe buy at each price? Perhaps, in the future I will alter the statement to read How many stamps did Joe buy at each price, given that at least one of each was bought? Your solution is perhaps shorter than mine as I did not make use of mod relationships. My answer is W = 78, X = 13, Y = 5, and Z = 4. Is the only solution. ??? My path wos:Joe came back from a stamp collectors gathering and told his sister, Jill, that he bought a hundred stamps. Joe said he bought the stamps at four different prices: \$0.59, \$1.99, \$2.87, and \$3.44 each. Jill asked, "How much did you pay altogether?" Joe replied, "One hundred dollars exactly." How many stamps did Joe buy at each price? Given: (1)--W + X + Y + Z = 100 and (2)--.59W + 1.99X + 2.87Y + 3.44Z = 100.1--Multiply (2) by 100--->59W + 199X + 287Y + 344Z = 10000. (3)2--Multiply (1) by 59----->59W + 59X + 59Y + 59Z = 5900. (4)3--Subtract (4) from (3)--->140X + 228Y + 285Z = 4100. (5)4--Divide (5) through by 140--->X + Y + 88Y/140 + 2Z + 5Z/140 = 29 + 40/140.5--Solving for X = 29 - Y - 2Z - (88Y + 5Z - 40)/140. (6)6--Set (88Y + 5Z - 40)/140 = u = an integer.7--Rearranging, 140u = 88Y + 5Z - 40. (7)8--Dividing (7) through by 5--->28u = 17Y + 3Y/5 + Z - 8. (8)9--Solving for Z, Z = 28u - 17Y - 3Y/5 + 8. (9)10--With 3Y/5 = integer, multiply by 2 yielding 6Y/5. (10)11--Dividing (10) out yields 6Y/5 = Y + Y/5 where Y/5 must be an integer also.12--Set Y/5 = v whence Y = 5v. (11)13--Substituting (11) into (8)--->Z = 28u - 85v - 3v + 8 = 28u - 88v + 8. (12)14--Substituting (11) and (12) into (6)--->X = 29 - 5v -56u + 176v -16 - (440v + 140u - 440v + 40 - 40)/140. (13)15--Simplifying (13)--->X = 29 - 5v -56u + 176v - 16 - u = 13 + 171v - 57u. (13)16--Substituting (11), (12), and (13) into (1)---> W + 13 + 171v - 57u + 5v + 28u - 88v + 8 = 100. (14)17--Simplifying (14)---> W = 29u - 88v + 79. (14)18--From (11) v =/> 1 and from (12) u =/> 3.19--Trying v = 1 and u = 3, W = 78, X = 13, Y = 5, and Z = 4.20--Checking--->78 + 13 + 5 + 4 = 100. Okay.21--Checking--->78(.59) + 13(1.99) + 5(2.87) + 4(3.44) = 46.02 + 25.87 + 14.35 + 13.76 = 100. Okay.22--Trying v = 2 and u = 3, X = 184 exceeding total of 100, therefore invalid.23--Trying v = 1 and u = 4, X = negative number, therefore invalid.24--All other values of u and v produce invalid results.25--Therefore W = 78, X = 13, Y = 5, and Z = 4 is the only solution. QED!

someguy

143 Posts

 Posted - 03/24/2013 :  23:56:01 Hi TchrWill, it is always nice to see alternate methods.I agree that you should modify the statement of the questionto say at least one stamp of each value was purchased, otherwisethere are 3 valid solutions.77, 13, 10, and 078, 13, 5, and 479, 13, 0, and 8I really should have checked the values D=0, D=4, D=8, and D=12as stated above in my final note rather than just assuming youwould ask a question with a unique solution. I should know better than to make an assumption like that.I look forward to your next one.
Topic
 New Topic  Reply to Topic  Printer Friendly Jump To: Select Forum New Visitor Forum       Testing Forum Homework Help Forums       Basic Math and Pre-Algebra       Algebra       Geometry and Trigonometry       Pre-Calculus and Calculus       Probability and Statistics       Standardized Test Preparation Help       Miscellaneous Math Topics Educator Forum       Teacher Talk Parent Forum       Parent's Place  -------------------- Home Active Topics Frequently Asked Questions Member Information Search Page
 Math Forums @ Math Goodies © 2000-2004 Snitz Communications