An old and fiendish puzzle this week. Last year I went to a New Year’s party. There were 8 children there and they all happened to be wearing black sweaters. My uncle gathered them together, found some chalk, and wrote one number on the back of each child. The numbers were: 1, 2, 3, 4, 5, 7, 8, 9. My uncle then asked whether it was possible to arrange the children into two groups of four, such that the numbers on the backs of the children in each group came to the same total. After much messing around, we managed it. Can you?
As ever, please do NOT post your answer, but feel free to say if you have solved it and how long it took. Answer on Monday. Oh, and happy new year.
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December 31, 2010 at 5:35 am |
FIRST!
3 minutes thinking outside the box (and it helps my father was a gymnast)
January 1, 2011 at 3:55 pm
Worked it out after reading your gymnast clue! Lovely!
January 3, 2011 at 5:31 am
Me too!! Awesome.
December 31, 2010 at 5:36 am |
I found an answer that I think satisfies the question in approximately 60 seconds.
December 31, 2010 at 5:37 am |
20 seconds for two creative solutions that seem like cheats, and they open up a whole family of similar cheats.
December 31, 2010 at 5:53 am |
Wrote numbers on post-it notes and I am embarrassed to say it then took me 3 minutes to figure a solution. Math, obviously, is not my strong point.
December 31, 2010 at 5:55 am |
About 30 seconds, although I think I may have cheated. There must be a certain amount of creativity required though.
December 31, 2010 at 5:56 am |
I think I got it.
December 31, 2010 at 6:20 am |
Got it in about a minute. As always, maths is only half the story, carefully reading and interpreting the question is the other.
Richard – Thanks for the fine puzzles and posts, and good luck for the New Year everyone!
December 31, 2010 at 6:33 am |
I think I solved it before I finished reading the question. Will wait for Richard’s answer to be sure.
December 31, 2010 at 6:52 am |
I got an answer in about two minutes, but I don’t like the feeling of having to cheat.
December 31, 2010 at 7:06 am |
Oooh, about 20 seconds, providing I’m right of course!
December 31, 2010 at 7:21 am |
About 10 seconds, assuming ‘messing about’ includes acrobatics.
December 31, 2010 at 7:23 am |
Two minutes today, going to take stickers to tonights party. Fun picking number 9….;-)
December 31, 2010 at 7:42 am |
First added the numbers noticing that six was missing got a result that was not divisible by two implying there was jiggery-pokery going on. My technique was to make number pairs and then you are faced with changing something. did and it worked.
Less than a minute in all
December 31, 2010 at 7:48 am |
Geez, how simple can you get? Took longer to read it than to solve it.
December 31, 2010 at 4:21 pm
Hah, wimp. I didn’t even need to begin reading it. I was born knowing the answer.
December 31, 2010 at 6:46 pm
Well I lost my memory but was hypnotized and went through memory regression and found I knew the answer in a past life……lol….
December 31, 2010 at 8:07 am |
1 minute and a handful of seconds – Looking forward to Monday to find if MY solution is in fact THE solution. Happy New Year one and all.
December 31, 2010 at 8:34 am |
Whoopeee one I can answer almost instantly.
December 31, 2010 at 8:53 am |
I won’t lie, it took me friggin’ forever. very entertaining puzzle.
December 31, 2010 at 9:04 am |
Ah got it with the “clue” from the first comment
December 31, 2010 at 12:06 pm
Same holds for me,
1 minute without a satisfiable solution;
after reading the first comment /clue another 10 sec for the real answer.
December 31, 2010 at 9:16 am |
Figured out almost immediately that it was impossible if taken entirely at face value, spent about a minute pondering the catch, then also clicked as soon as I read Todio’s comment.
December 31, 2010 at 9:49 am |
I came up with an answer almost immediately but it might only work for a short period of time
December 31, 2010 at 9:54 am |
The numbers as they are add up to an odd number and hence it is not possible to divide their total as they are. That means, the solution lies in some kind of ‘cheating’ where some number is to be interpreted as some other number. If this strategy is acceptable, there would be n number of solutions.
December 31, 2010 at 10:16 am |
I so hate to say this but I used a whole sheet of paper….yesss both sides and half a pencil ….got mad through ball of paper…. ceiling fan through it back at me…..hit me in the head too….serves me right I guess….had a snack…..good apple too….then looked again….RE-READ and duhhhh…smack head and got it…..got to be right cause I can’t think of any other way…..ohhhh wait….there might be some other ways if you get really, really mathematically monkeying around…..but no don’t think all of that is necessary…..might be fun but simple may be best….. all in all total with snack time included….20 minutes…..hang head in shame……. :}
December 31, 2010 at 9:50 pm
Well I feel better about myself now…just found 3 combinations of 2 equal groups of 4 …… and IF it doesn’t matter what we do to the children in those 2 groups of 4 there maybe many ways to make them into equal equations……he he he …. anyone have a sharp saw and heavy duty needle and thread…… that way no one has to die*
* for any first timers here Richard {our wonderful host} is good about posting Friday Puzzles that result in the untimely demise of someone or something…..lol…but all done in good taste mind you…… he he he..
December 31, 2010 at 11:40 am |
Took me a good 5 minutes, I feel quite silly after seeing all these other people who claim to have done it in 20 seconds.
First comment helped quite a lot
December 31, 2010 at 11:41 am |
i got a solution in few seconds
assuming 8 children numbered 1,2,3,4,5,6,7,8. because u added 9 there?
anyway 9 can’t be grouped into 2 groups of 4 lol
December 31, 2010 at 11:44 am |
ohh 6 is missing lol ok ok i got it lol i thougth it was a typo
December 31, 2010 at 12:15 pm |
I got the answer straight away, but it took me about 30 minutes to round up enough willing children to prove it..
December 31, 2010 at 12:21 pm |
For once it took me about ten seconds. They usually completely stump me.
Probably a fluke :p
December 31, 2010 at 12:22 pm |
Got it almost straight away, because there was obviously something amiss.
December 31, 2010 at 1:25 pm |
Got it in 30 seconds.
December 31, 2010 at 2:00 pm |
I got the answerrrrrr!!! Took me like 5 mins. lol
December 31, 2010 at 2:18 pm |
10 seconds deciding it was impossible. 1 minute to think of a way to cheat. 5 minutes to come up with a cheating answer.
December 31, 2010 at 3:04 pm |
“Cheat” is such a harsh word. I prefer “thinking outside the box”. And this is more like it – a challenge that isn’t simply algebra. Took me about 10 seconds, but only because I saw the cheat^H^H^H^H^Htwist right away.
December 31, 2010 at 3:15 pm |
erm. 2 mins or so. i dint read the question properly. but i thought the puzzle was a cheap trick cuz if i saw ur granpa writing or typig then i wud have surely noticed the missing 6!
December 31, 2010 at 4:08 pm |
nearly instantaneous. helped that i knew some entertaining legends about gauss and mobius.
December 31, 2010 at 5:18 pm |
Think I solved it with a strange move in 3 min.
December 31, 2010 at 6:17 pm |
The “Calendar Cubes” trick, less disguissed, I’d say.
“A man has two cubes on his desk. every day he arranges both cubes so that the front faces show the current day of the month (from 01 to 31). What numbers are on the faces of the cubes to allow this?”
December 31, 2010 at 7:14 pm |
Took about a minute.
December 31, 2010 at 7:36 pm |
Looks like the time is moving on, so to everyone in a time zone east of me and across the pond as you will bring in the New Year of 2011 before me….let me now wish you a healthy, safe and happy new year! Richard thank you so much for the Friday Puzzles. I enjoy them and hope to do so in 2011!
January 1, 2011 at 12:11 am |
[...] It’s the Friday Puzzle! An old and fiendish puzzle this week. Last year I went to a New Year’s party. There were 8 children there and they [...] [...]
January 1, 2011 at 12:58 am |
I thought of two possible answers in about 5 minutes.
January 1, 2011 at 3:44 am |
Pretty quickly after reading Steve’s clue! (:-))
January 1, 2011 at 6:35 am |
Its missing 6 so it slits up evenly into 2 groups. 10 seconds
January 1, 2011 at 6:40 am |
sorry more like 3
January 1, 2011 at 8:17 am |
20 seconds, too easy.
January 1, 2011 at 1:55 pm |
Yes, about 30 seconds. Like it!
January 1, 2011 at 3:34 pm |
This one was surprisingly easy for me
10 seconds or something like that.
January 1, 2011 at 5:14 pm |
There are four pairs and two pairs in the group makes equal totals.
January 1, 2011 at 6:46 pm |
It can’t be done without cheating (e.g. changing one number into another.)
January 1, 2011 at 8:51 pm |
About 4 minutes
January 1, 2011 at 10:34 pm |
well for me i didn’t find the answer to the puzzle as it is bcoz my brain resisted the idea that 39/2 = a whole number??? so i decided to use the chalk lol
January 2, 2011 at 12:36 pm |
ca 1 min reaching _a_ solution. Another minute or so to accept that this i probably _the_ solution.
January 2, 2011 at 2:56 pm |
I think I actually got this one. Took about 30 seconds.
January 2, 2011 at 5:07 pm |
Two solutions in three minutes.
January 2, 2011 at 9:35 pm |
less then a minute for one solution…
few seconds more for another one
February 4, 2011 at 5:47 am |
First thought was to sum it up.
In Python:
>>> sum([1, 2, 3, 4, 5, 7, 8, 9])
39
# Hrm, obviously it’s a trick. Let’s get something even…
>>> sum([12, 3, 4, 5, 7, 8, 9])
48
# So 12 + x + y = 48/2 = 12
12,3,9 ; 4,5,7,8
or
12,4,8 ; 3,5,7,9
or
12,5,7 ; 3,4,8,9