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?

Enjoy

I 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

3 L A T E S T R E P L I E S (Newest First)

someguy

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 question to say at least one stamp of each value was purchased, otherwise there are 3 valid solutions.

77, 13, 10, and 0

78, 13, 5, and 4

79, 13, 0, and 8

I really should have checked the values D=0, D=4, D=8, and D=12 as stated above in my final note rather than just assuming you would 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.

TchrWill

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

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 = 100 59A+199B+287C+344D=10000

Substituting A=100-B-C-D into the second equation gives

140B + 228C + 285D = 4100

Note 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 3 7B is congruent to -4 mod 19

This means B = 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 to

B=13 228C + 285D = 2280

We 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=100 77(.59) + 13(1.99) + 10(2.87) + 0(3.44) = 100.00

Note: if we had not 'spotted the solution', we could have used the last equation involving 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 $100 lets 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.