- Three couples bought candy at a candy store. Each person paid as many cents per candy as pieces of candy he or she bought, and everyone bought a different number of pieces (no one was empty handed.) When they compared their bills, they found that the difference between the amounts paid by each husband and wife was the same. What is the smallest number of candies purchased? Show the amounts for the difference and for each person/couple.
- What if there were two couples?
- Four couples?
- Five couples?
- Can a method be extended for M couples?
- If the difference between each husband's and wife's amount is less than 500 cents, what was the largest number of couples possible in the store?

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Guest Message by DevFuse

# Couple Candy Shopping

Started by BMAD, Dec 13 2013 12:39 AM

2 replies to this topic

### #1

Posted 13 December 2013 - 12:39 AM

### #2

Posted 13 December 2013 - 07:43 AM

Spoiler for

### #3

Posted 30 December 2013 - 08:02 PM

Nicely done Rainman!

The break through was coming up with (x + y) (x - y)!

I dabbled with variations on x^{n} but found that too many of the combinations yielded a zero. So I came to the same conclusion that you had to add another prime number to the mix and that 3 yielded the lowest results.

Interesting puzzle, nice elegant solution.

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