Comments (21)

1. Good early start for the brain!
   You should probably also indicate that the 5 candidates were also allowed to vote. I ran under the assumption that there were 26 votes.

   posted by J on 11-3-2009 at 8:16 am

2. Too easy. Could have made it a school wide election to make the numbers bigger.

   Not complaining though, I love these things.

   posted by John L. on 11-3-2009 at 8:22 am

3. This was very nice. I would be interested to know how you arrived at your solution. I spent an inordinate amount of time this morning trying to work this out with algebra. I finally had to plot y=31-2x to find a workable solution.

   posted by Josh on 11-3-2009 at 8:26 am

4. I just picked an even number I knew was divisible by 2, 3, and 6 and came up with 12 pretty quickly, which then gave me 6, 4, and 2 votes for the others with 7 left over for the last person.

   posted by John L. on 11-3-2009 at 8:31 am

5. Josh,

   at least I arrived to conclusion very simply by assuming the least amount of votes that satisfied the conditions. (This was easy as we knew there weren’t very many votes overall.)

   Starting with Woody having 1 vote, one notes that we would have at least two people with uneven votes. Second smallest guess was Woody’s 2 votes. When we get 4 even numbers for these people and see that we are left with 7 for Willy, we have the answer. This was also probably the fastest way to get the answer: it basically gives the answer as fast as you can calculate the sum of the votes.

   posted by kl on 11-3-2009 at 8:32 am

6. There is another solution, although it is a bit of a cop-out: Willy wins unanimously (31 votes).

   posted by Scott on 11-3-2009 at 8:40 am

7. John, I followed the same logic. I just guessed and started with 12. Everything else surprisingly worked out from there.

   posted by K on 11-3-2009 at 8:55 am

8. @Josh : I did a slightly longer calculation than John –

   Wally+ Wendy+ Willy+ Woody+ Wyatt = 31
   Wyatt=2(wendy) , Wyatt=3(wally) , Wyatt=6(woddy)

   wyatt/3 + wyatt/2 + Willy + Wyatt / 6 + wyatt = 31
   wyatt= (31 – willy) /2

   Willy received an odd number of votes.

   If Willy received 1 vote rest of the votes would look like –
   Willy Watt Wendy wally Woody
   1 15 7.5 5 2.5
   ( not possible – no fractions allowed)

   If Willy received 3 votes rest of the votes would look like –
   Willy Watt Wendy wally Woody
   3 14 7 4.666666667 2.333333333

   And so on. ( I cheated and used excel here! )

   the only 2 options with no fractions were –
   Willy Watt Wendy wally Woody
   7 12 6 4 2

   And

   Willy Watt Wendy wally Woody
   19 6 3 2 1

   And the only set of numbers that fit are

   Willy Watt Wendy wally Woody
   7 12 6 4 2

   HTH.

   posted by Venky V on 11-3-2009 at 9:17 am

9. J, it did say that everyone voted. It did not mention that the candidates do not vote. In a regular election, the candidates still vote.

   posted by Ryan on 11-3-2009 at 9:43 am

10. How about this:
    Wyatt had 20 votes
    willy had 4 votes
    Wendy had 4 votes
    Wally had 2 votes
    Woody had 1 vote

    posted by Christine on 11-3-2009 at 9:52 am

11. Another highly improbable solution would be for Willy to receive 31 one votes and all other candidates to receive 0. Of course that means the other candidates didn’t vote for themselves.

    posted by Jon on 11-3-2009 at 9:55 am

12. I got the above solution, but there’s an alternate:

    Wally: 0
    Wendy: 0
    Willy: 31
    Woody: 0
    Wyatt: 0

    posted by SP on 11-3-2009 at 10:10 am

13. I started with algebra, but once it dawned on me that you cant have partial vote, gridding it was way easier

    posted by Rod on 11-3-2009 at 10:53 am

14. I’m not sure where others are getting it, but the solution 31 votes for Wyatt and 0 votes for all other candidates is not possible. The description states that Wyatt’s vote count is equal to 2 times, 3 times, and 6 times the votes of Wendy, Wally, and Woody respectively. As 0 times any number is still 0, the 31-0-0-0-0 solution is not correct.

    posted by Ryan Smith on 11-3-2009 at 11:05 am

15. Silly me, I misread others’ solutions, previous comment withdrawn :-)

    posted by Ryan Smith on 11-3-2009 at 11:08 am

16. Christine, I’m afraid that answer doesn’t quite work.

    Remember in the clues that Wyatt earned twice as many votes as Wendy.

    Your solution has Wyatt with 20 and Wendy with 4, which would be five times as many.

    posted by Sandy Wood on 11-3-2009 at 11:16 am

17. willy:7
    wyatt:12
    wendy:6
    wally:4
    woody:2

    posted by sam on 11-3-2009 at 11:53 am

18. Great puzzle! Keeping the names/variables straight was almost more challenging than answering the question. :)

    posted by HeyBeckyJ on 11-3-2009 at 12:56 pm

19. I don’t like trial and error with all the arithmetic, so my key was to recognize that Willy must be the odd vote total because:

    Given: Wyatt=x, Willy=y

    x + (3/6)x + (2/6)x + (1/6)x + y = 31

    reduces to:

    2x + y = 31

    2 times any number is even, so Willy (y) must be odd to get an odd sum (31).

    And Wyatt (x) and must be a multiple of 6, and must be less than 15.

    So Wyatt must be 12.

    posted by Larchie on 11-3-2009 at 1:50 pm

20. To keep the variables straight I used the second letter of their names. Wyatt = Y, Wendy = E…. on to A, O, and I.

    It was pretty easy; once I saw that Y = 6*O my “order of magnitude” sense told me that O was going to be either 1 or 2.

    posted by Andrew on 11-3-2009 at 3:09 pm

21. Andrew, nice call on the variables – I did the same exact thing when devising the puzzle.

    posted by Sandy Wood on 11-3-2009 at 9:27 pm

Comment