All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Numbers
Pennies, Dimes, and Quarters (Posted on 2009-02-17) Difficulty: 2 of 5
In my left pocket I have a mix of pennies (1 cent) and quarters (25 cents). In my right pocket I have a bunch of dimes (10 cents). The number of coins in each pocket is the same, so is the cash value. What is the smallest (nonzero) number of coins I can have?

What if I had coins of x cents and z cents in my left pocket and coins of y cents in my right pocket - is there some quantitiy of coins I can have so that each pocket has the same number of coins and same cash value? (The value of the coins are positive integers x > y > z.)

See The Solution Submitted by Brian Smith    
Rating: 3.5000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts On the generalization | Comment 7 of 11 |
This is really a beautiful symmetrical result.  The other posts show the algebra, but allow me to show how simple it becomes in practice:

The number of each type of coins is just the difference of the values of the other two (provided you then divide out the gcd of these differences.)

For values 1, 25, 10 The differences:
25-10=15 pennies
10-1=9 quarters
25-1=24 dimes
but since 15, 9 and 24 are all divisible by 3 we get

5 pennies, 3 quarters, 24 dimes.



  Posted by Jer on 2009-02-17 14:39:33
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information