On Friday I set this puzzle….

You have nine tennis balls and four shopping bags. Your challenge is to put the balls in the bags in such a way that there is an odd number of balls in each bag. That is, each bag must contain 1, 3, 5, 7 or 9 balls. Can you do it?

If you haven’t tried to solve it, have a go now. For everyone else the answer is after the break….

It is all about putting the bags inside one another. You could, for example, put 3 balls in each of 3 bags, and then place one of these bags in the fourth bag. There are lots of possibilities. Did you solve it? Any other solutions?

I have produced an ebook containing 101 of the previous Friday Puzzles! It is called **PUZZLED** and is available for the **Kindle** (UK here and USA here) and on the **iBookstore** (UK here in the USA here). You can try 101 of the puzzles for free here.

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I put all 9 balls in one bag, and put all four bags inside each other. I got the answer in about half a second, and didn’t bother working out how many other possible combinations there are.

Yes, that was my answer after a couple of minutes.

Nice.

Damn. I spent too much time proving to myself that you can’t split an odd number into four other odd numbers😦

Here are my solutions, there are 69 of them, and looking at them again I’m pretty sure there are a at least a few more that I missed.

((((9))))

(8(((1))))

((8((1))))

(((8(1))))

(6(((3))))

((6((3))))

(((6(3))))

(6(2((1))))

(6((2(1))))

((6(2(1))))

(4(((5))))

((4((5))))

(((4(5))))

(4(2((3))))

(4((2(3))))

((4(2(3))))

(4(4((1))))

(4((4(1))))

((4(4(1))))

(4(2(2(1))))

(2(((7))))

((2((7))))

(((2(7))))

(2(6((1))))

(2((6(1))))

((2(6(1))))

(2(4((3))))

(2((4(3))))

((2(4(3))))

(2(2((5))))

(2((2(5))))

((2(2(5))))

(2(2(2(3))))

(6(1)(1)(1))

(4(3)(1)(1))

(2(3)(3)(1))

(2(5)(1)(1))

((3)(3)(3))

(7((1))(1))

(5((3))(1))

(5(2(1))(1))

(5((1))(3))

(3((5))(1))

(3(2(3))(1))

(3(4(1))(1))

(3((3))(3))

(3(2(1))(3))

(3((1))(5))

(1((7))(1))

(1(2(5))(1))

(1(4(3))(1))

(1(6(1))(1))

(1(4(1))(3))

(1(2(1))(5))

(1((1))(7))

(1(2(3))(3))

(1)(1)((7))

(1)(1)(2(5))

(1)(1)(4(3))

(1)(1)(6(1))

(1)(3)((5))

(1)(3)(2(3))

(1)(3)(4(1))

(1)(5)((3))

(1)(5)(2(1))

(1)(7)((1))

(3)(3)((3))

(3)(3)(2(1))

(3)(5)((1))

Your solutions include even numbers in bags. The article specifies odd numbers only

Chickenhawk

I think you are misinterpretting

Emyln’s terminology

Each matched pair of parentheses represents a bag.

Bags should only have an even number of balls in them directly if they also have an odd number of balls inside of other bags, so that the total is odd. If there are any solutions above where that is not the case then they are errors

Chicken little is that you?

I’m counting 128,

http://pastebin.com/GFPWA1bW

I think I zapped any duplicates.

You could also just put one / three tennis balls in each bag, and throw the rest away. The question doesn’t specify that ALL the balls have to go into the bags.

Cheers, Richard Hope the weekend went well and proved fascinating, entertaining and delightfully successful. Yep, I came up with a few versions of the answer, the first and immediate being to put one bag inside another, until all of the bags were inside one, forming, essentially, one multi-layered bag and then placing all nine balls into that one bag; so, yep, essentially the same answer. Thanks for another good one

Nice puzzle . Great to see you last week in Edinburgh – loved your psychobabble talk.

An ultimately pointless puzzle – and yes I know many of them are, but at least some have a use as an exercise in lateral thinking. This puzzle gives you no help at all. Okay – so we’re feeling smug that we’ve put balls inside a bag inside a bag inside a bag…..huzzah! Until the real world guy looks at you and says “why have you quadruple bagged those balls, but wasteful isn’t it?” or “yeah…but we need the bags on different courts in the sports centre – now what are we supposed to do?”

Bah Humbug !

There _must_ be an odd number of balls in each bag? Sounds a little OCD to me.

Around a minute I guess. Bag inside a bag was my main solution. Only using 8 balls (5+1+1+1) was my backup.

Once you realise that the sum of an even number of odd numbers can never equal an odd number (4 bags with odd numbers, sum equal 9), it becomes obvious, that balls somehow have to either not be counted (thrown away) or counted more than once (bag inside bag).

“in each bag” – not in bags within bags or throw away all the bags you want- I knew it couldn’t be a literal answer and be correct so I figured it wwas yet another cheat or trick, non-literal answer and didn’t bother.

If you look at each bag in turn and count the number of balls inside it, and that number is odd, that is a valid solution. I don’t see why that would be cheating.

Wait a minute: a bag is not a ball and 0 is not odd… If each bag must contain 1, 3, 5, 7 or 9 balls and the previous id true, there is no solution..

No a bag is not a ball. Nor is 0 odd. I don’t see how either of those claims invalidates Richard’s answer.

The point is you put three balls into bag A. Now put bag A inside bag B and bag B also contains an odd number of balls (3). Nothing requires that a bag be a ball.

First of all: a question is a question is a question. Second what is the unit of interest here? Thanks for restating that a bag is not a ball and zero is not odd, repeating rhetorical statements always helps.

Anyway, bag B in your example contains bag A and not 3 balls.

If I have a useful descriptive unit (like bags) to state the content of a container (like a bag in a bag) why do an iterative push? The problem clearly states that each bag must contain balls and also specifies the number of balls (1, 3, 5, 7 or 9). Bag B (in your example) doesn’t contain any balls, it contains bag A. It could though, if you put 2 balls in bag A, 1 ball in bag B and bag A in bag B, but still this wouldn’t make bag B contain 3 balls, just one ball and a bag, we are talking about contents of bags after all…

Niccolo,

Your argument is like saying ” 5 people are in a room, but 4 of them are wearing clothes. Since 4 people are inside their clothes, that means only one person is inside the room”.

Nonsense, of course.

Jiminy,

You would be correct if we also said that balls wear bags… but of course we don’t, that would be stupid.

Niccolò,

A bag contains *everything* within it, even if something are contained in another object.

“Bag B (in your example) doesn’t contain any balls, it contains bag A”

The balls within bag A are also contained within Bag B … The total contents of Bag B are one bag and 3 balls. You seem to be missing that point. If I have a closet containing 3 shoe boxes each containing a pair of shoes and you were asked how many shoes were in the closet, your answer would be 0 ? Somehow you think the shoes wouldn’t count as being in the closet because they were also contained within some other object ?

Stu,

yes I would if the closet contained other closets. If I had a closet with 3 closets inside and inside those 3 pais of shoes (is that even possible :P) I wouldn’t say meta-closet has no shoes I would ask which closet, being that meta-closet would not have shoes in it specifically only sub-closet would. The location of closets or bags is independent from their content. My problem is not lack of understanding of the answer, I got it was the only possible answer given the problem. My problem is with how the question was stated. The variables we are asked to consider are bags, balls and content. Considering bags if I have 2 nested bags the content of the meta bag is another bag and not the last content of the sub-bags…. I just feel the way the question is posed is disingenuous, then again it is a puzzle and trick questions are always a possibility…

They must love you at airport security. When they ask you if you have any liquids in your bag you say no, because your bag contains only contains a bottle and it is the bottle that contains the water not the bag. That is just a misunderstanding of what the wonder contains means.

What constitutes another container in your world view.. if something is completely surrounded by air within your bag (i.e your bag contains one ball and as you are walking it bounces around in the bag) would it still not be in the bag ? It’s in the air, and it is the air that’s in your bag… Or would it be switching between being in your bag and not in your bag when it touched a part of the bag as apposed to when it was bouncing and the air was around it. If you don’t consider the molecules in air to be another container, what is your criteria, a solid object (how solid? would it be in your bag if it was contained in a string mesh), would you consider it in your bag if it was contained in an open bag ? Wrapped in cloth ? Partially wrapped in cloth ? Would it be a Schrödinger’s ball, and you would have to open your bag and determine the state of the contained bag to determine whether or not your bag had a ball in ?

Just for correctness, airport security asks if YOU are carrying liquids or other bla bla blah… not for the content of your bag. But if they did ask for the content of your bag and you had 2, they would have to ask twice and qualify which bag they are talking about! One might not have liquids and the other one might have them. If one bag is inside the other is irrelevant.

I don’t know how much you will follow of the following but this is how I see the “problem”. I chose python syntax because its the most immediately understandable. Btw the “problem” is actually no problem because apparently everyone (including me) understood the question and the possible solutions. However:

What is in the bag?

def search(container):

for obj in container:

open container

print obj

#This is what I would do. What everyone is saying is that I should make it recursive like adding #something like:

try:

search(obj)

except:

pass

This would mean that if someone asked you what’s in the bag and the bag contains a wallet, one would have to answer: A wallet plus all of the wallet’s contents.

I didn’t quite get your point about air and cloth or something, but I want to make a similar one: Imagine you have a bag with a phone, a wallet and another bag in it. Now imagine you have N objects in the wallet and N^2 objects in the bag. The question is the same: What’s in the bag? How big would N have to be for you to answer: a phone, a wallet and a bag?

Would you always also answer all of the components of the phone like battery, screen, liquid crystals? Do you have to enumerate all the crystals one by one or just saying screen is sufficient? What about the stitching of the wallet? is the thread considered also as being inside the bag or the wallet or both? But why stop there, stuff is made of atoms, do I have to atomize all the content of the bag independently of what’s in it? What about subatomic particles? My first comment had a question in it, what’s the unit that you employ and what are the rules to change the unit? or similar…

Anyway, I grow tired.

I give up…

One ball in each bag is the easiest solution, since the wording does not require that we bag all the balls. Given the triviality of some of the recent puzzles we have had and lack of care over wordings, I presumed that the solution to this one would either be the word “no” or the answer I gave. I didn’t even bother to investigate whether there might be a clever solution. Come on Richard, try a bit harder to communicate the problem you want to set.

He does use the definite article which I would infer to mean all the balls…

Come on Ivan, try a bit harder solve the problem Richard has set.

Lol

It amuses me how many people in recent weeks have blamed Richard’s wording of the question, for their own failure to get the answer.

I just read the puzzle, guessed the cheat,

and am now forcing the bags into a ball to see how many will fit in

I put one bag inside the other, like Russian stacking dolls, and put all the balls inside the top bag. I realized there are many other similar possibilities. This is a good example of thinking inside the box (or bag)!

If one of the balls is a girl, what are the chances the other eight are girls? Just making trouble…

Need better puzzles here, boss.

I just put 1 ball in each bag. It was never stated you had to use all the balls.

The problem states, “…put THE balls in THE bags…”. That implies “THE” balls that you have, all nine. If the problem stated, “..put balls in bags…”, you could use however many, or few, you wished. Just saying.

The puzzle didn’t specify all the balls. I just had 5 left loose!

Thanks Richard. my answer was (2(1)) (3) (3).

(1(1(1(1+1+1)1)1)1) to keep things symmetrical I have kept the balls apart in each bag.

Tedious

Very clever. I never would have thought of that one!

I am a different “Jerry” from the one several comments above

There’s a similar puzzle in Richard’s book, so I knew the answer immediately.

TWO SOLUTIONS WITH NO NESTING OF BAGS

1. Put one ball in each of the bags. Put the rest of the balls in any bag(s) you like. Now, each bag has one ball (even though some have more). One is an odd number. Done!

2. Put one ball in each bag. Leave the rest out. Nowhere did the puzzle say you have to use all the balls.

This is similar to the question, “How many months have 30 days?” Answer: Eleven; February is the only exception.

Oops. I just noticed earlier replies had my second answer. Sorry.

The people who claim that the puzzle does not specify that all nine balls have to be put inside the four bags, are the people who do not know what “the” means.

They know what Loony means.

I first thought, you could put all 9 balls in one bag, then put that bag in the other two. I then though, nope, that can’t be the answer -_-

*other three

Ballbags.

That’s all I have to say on this matter.

how to arrange the 90 people in nine room as the order of odd no.

plz giv the solution .

how to arrange the 90 people in nine room as the order of odd no.

There are 90 horses and 9 rooms.. But the horses should be divided by odd number in each room… nd no room should remain empty

how many meiotic divisions are needed for producing 1000 zygotes?’

how many meiotic divisions are needed for producing 1000 zygotes?’

50 Bold and 5 box feel in every odd no.