testing header
Math Goodies is a free math help portal for students, teachers, and parents.
Free Math
Newsletter
 
 
Interactive Math Goodies Software

Buy Math Goodies Software
testing left nav
Math Forums @ Math Goodies
Math Forums @ Math Goodies
Home | Profile | Register | Active Topics | Members | Search | FAQ
Username:
Password:
Save Password
Forgot your Password?

 All Forums
 Homework Help Forums
 Miscellaneous Math Topics
 Stamp Collector
 New Topic  Reply to Topic
 Printer Friendly
Author Previous Topic Topic Next Topic  

TchrWill
Advanced Member

USA
80 Posts

Posted - 03/24/2013 :  08:06:48  Show Profile  Reply with Quote
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
Go to Top of Page

someguy
Advanced Member

Canada
143 Posts

Posted - 03/24/2013 :  17:48:50  Show Profile  Reply with Quote
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.
Go to Top of Page

TchrWill
Advanced Member

USA
80 Posts

Posted - 03/24/2013 :  22:23:25  Show Profile  Reply with Quote
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!
Go to Top of Page

someguy
Advanced Member

Canada
143 Posts

Posted - 03/24/2013 :  23:56:01  Show Profile  Reply with Quote
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.
Go to Top of Page
  Previous Topic Topic Next Topic  
 New Topic  Reply to Topic
 Printer Friendly
Jump To:
Math Forums @ Math Goodies © 2000-2004 Snitz Communications Go To Top Of Page
This page was generated in 0.09 seconds. Snitz Forums 2000
testing footer
About Us | Contact Us | Advertise with Us | Facebook | Blog | Recommend This Page




Copyright © 1998-2014 Mrs. Glosser's Math Goodies. All Rights Reserved.

A Hotchalk/Glam Partner Site - Last Modified 21 May 2014