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November 22

4th grade stumper.

I work in a 4th grade classroom. Anyways, a math problem that the class worked on in groups, was: If you have 10 socks, 5 black and 5 blue. Its dark in your room so you cant see. What is the minimum amount of socks you would need to pull out to ensure you had a match.

Now, the majority of the students stated it was 3. However, the book states that the correct answer is 6. I happen to agree with the students in that i think it is 3. Can anyone shed some logic on the solution of 6? Or put it to rest? I need to save some face tomorrow!!

thanks72.74.29.159 01:48, 22 November 2006 (UTC)moe.ron[reply]

As far as I can tell, 3 is the correct answer. Perhaps they were thinking of the related question: "How many socks would one have to pull out to insure you have at least one of each color?" The answer to that, somewhat less practical question, is 6. -GTBacchus^(talk) 01:51, 22 November 2006 (UTC)[reply]

Just a comment, I go to school and books are very frequently wrong, especially math books. (not that frequently but it happens at least once per book...) Cbrown1023 01:56, 22 November 2006 (UTC)[reply]

The correct answer is 3 by the pigeonhole principle: the holes are the colours, the socks are the pigeons, and more than one pigeon in a hole means two or more socks of the same colour. If you're going to a dress-differently party and the question had been how many socks you need to pull out to ensure you can go the party wearing mismatched socks, then 6 is your answer. --Lambiam ^Talk 03:25, 22 November 2006 (UTC)[reply]

Yeah, I remember spending a whole evening on a problem one without being able to get the solution in the book only to find out the next day that the book was wrong! Vespine 04:01, 22 November 2006 (UTC)[reply]

Another possible question is "how many socks do you need to take to ensure that you have a pair of BLUE socks". There the answer is 7, as it would be if you wanted to ensure you had a pair of black socks. StuRat 07:15, 22 November 2006 (UTC)[reply]

When I was in primary school, we were studying basic probability, and a homework question we had was "if you toss a fair coin 10 times and get heads every time, what will probably happen on the next toss?" with options of probably heads, probably tails, or even chances. I said even chances and the teacher told me I was wrong. I was quite unhappy about that one. Even supposed authorities can be wrong! Maelin (Talk | Contribs) 11:05, 22 November 2006 (UTC)[reply]

I sure hope you told your parents and had them take it up with the teacher and principal. Otherwise, that teacher may have continued to teach garbage for years. StuRat 18:46, 22 November 2006 (UTC)[reply]

In that teacher's defense, I will say that if we don't know for certain that the coin is fair (which is the situation in practice), then after 10 consecutive heads, the correct answer for the next toss is "probably heads". Probably not what he had in mind, but still. Just curious, what did he say was the probability for heads in the next toss? -- Meni Rosenfeld (talk) 20:20, 22 November 2006 (UTC)[reply]

She said it was tails. *shudders*. Maelin (Talk | Contribs) 10:58, 23 November 2006 (UTC)[reply]

It is probably in your best interests that you didn't take her up on it – you may have ended up with something like this sent home. :-) – AlbinoMonkey ^(Talk) 23:14, 23 November 2006 (UTC)[reply]

Well, 6 could be the answer if you have a strange idea of fashion like Dobby from the Harry Potter books. – b_jonas 18:02, 25 November 2006 (UTC)[reply]

Yeah, I figured she'd say tails was more probable, but did she mention some numeric probability? Or was that beyond her mathematical skills? It would have been a laugh if she had said "1 / 2048 probability of heads"... -- Meni Rosenfeld (talk) 21:22, 25 November 2006 (UTC)[reply]

6 only works as the answer if the question is "what is the minimum amount of socks you would need to pull to ensure you had a blue sock?" JPG-GR 21:50, 28 November 2006 (UTC)[reply]

How to find a pattern

Is there a method to finding the pattern in any set of numbers? I ask this question as an 11th precalculus student attempting to help my 11 year old cousin with her homeowork, the specific set of numbers she was given were (Solve x):

5  6  7  9  8
52 63 94 18 x

They are studying prime and composite numbers and I was dumbfounded when neither me nor my father could solve this problem. Any help, tips, or formulas would be greatly appreciated.—The preceding unsigned comment was added by 68.37.113.54 (talk • contribs) .

Is the placement of the 8 column at the end specified that way in the problem, or were the 9 and 8 cols. reversed? I see no obvious pattern. The placement of the 8 at the end seems to indicate to me that there may be more than a top row-bottom row linking, but I don't see anything there either, at least immediately. --TeaDrinker 03:16, 22 November 2006 (UTC)[reply]

Hint. It is indeed a relation between the top number and bottom number. Think of squaring. -- Jitse Niesen (talk) 03:27, 22 November 2006 (UTC)[reply]

You can actually find any number of patterns in a sequence of numbers. In such puzzles (I don't consider this mathematics) there is usually a particularly simple pattern, but I've seen cases in which the "official" solution was more complicated than another applicable pattern. --Lambiam ^Talk 03:43, 22 November 2006 (UTC)[reply]

If this is a mathematics problem it's a pretty crappy one, it doesn't actually use much mathematics as much as maybe lateral thinking. I guess it doesn't make it an invalid problem, but I can't see any possible application of this type of solution what so ever. If you need a hint there is no reason for the 8 to be after the 9, it could read 7>94, 8>x, 9>18, apart from that it's purely a puzzle for puzzle's sake.. Vespine 03:57, 22 November 2006 (UTC)[reply]

Reply - ability to solve this problem is related to the ability to crack codes - not something I'm interested in either.87.102.36.82 13:57, 22 November 2006 (UTC)[reply]

Smacks head and stops calculating in modular arithematic Oh yes, squaring does nicely. --TeaDrinker 04:08, 22 November 2006 (UTC)[reply]

Well, mod 9 there's a pattern you might have noticed, which might then have let to the real answer... Tesseran 05:31, 22 November 2006 (UTC)[reply]

Don't get how mod 9 helps here? Can't see any pattern in the remainders on division by 9?

For an arithmetical solution that works between 10 and 99 - how about using the sawtooth function to give effectively int(a/10) and (a-(10*int(a/10)) - and using that to swap the first and second digits...87.102.36.82 14:28, 22 November 2006 (UTC)[reply]

Sometimes my method of finding a pattern in numbers is to type them into OEIS (link), :P. Not that that would help you in this case, but maybe in the future. - Rainwarrior 17:09, 22 November 2006 (UTC)[reply]

Actually, searching OEIS for "52,63,94" does work, surprisingly enough. --Lambiam ^Talk 20:20, 22 November 2006 (UTC)[reply]

Ha ha ha! OEIS wins again! - Rainwarrior 16:45, 23 November 2006 (UTC)[reply]

For the record 10n²-99int(n²/10) where int(x) = x − sawtooth_function(2πx) will work.. Sorry87.102.36.82 19:18, 22 November 2006 (UTC)[reply]

If y=n²

Using L=1+int(log₁₀y)

Z_n=

\sum _{x=1}^{x=L}

10^x-1[int(y/10^L-x)-10int(y/10^L-x+1) ]

Giving the series

n    1, 2, 3, 4,  5,  6,  7,  8,  9, 10, 11,  12 etc
Z_n   1, 4, 9, 61, 52, 63, 94, 46, 18, 1, 121, 441 etc

too much time on hands...

OH MAN!! I'm wetting myself imagining the look on the teacher's face as the 11 year old shows them the above answer. Thoroughly impressive! Bravo. Vespine 22:10, 22 November 2006 (UTC)[reply]

By distributing the summation over the subtraction in 87.102.36.82's formula, substituting for the summation variables to bring terms with equal arguments to the int function together, and some further elementary algebraic manipulation, you can obtain this cryptic version:

Let y = n², and put L = 1 + int(log₁₀y)

Then Z_n = 10^L+1y – 99

\sum _{x=1}^{x=L}

10^L−xint(y/10^x–1).

For example, if n = 7, we get y = 49, L = 2, Z₇ = 1000×49 − 99(10×49 + 4) = 1000×49 − 99×494 = 49000 − 48906 = 94. --Lambiam ^Talk 04:31, 24 November 2006 (UTC)[reply]

Plotting this graph yields a very screwed up image. It doesn't look like any relationship that I've seen before.

Shall we call this the Wikipedia Reference Desk function? I would suggest "W function", but that's taken already. —AySz88 \^-^ 03:07, 25 November 2006 (UTC)[reply]

What is wrong.

Okay, let's see what you got!

What's wrong here:

$-1=i^{2}=i\times i={\sqrt {-1}}\times {\sqrt {-1}}={\sqrt {-1\times -1}}={\sqrt {1}}=1$

PureRumble 23:00, 22 November 2006 (UTC)

Fourth equal sign.--gwaihir 23:16, 22 November 2006 (UTC)[reply]

So the rule sqrt(a)*sqrt(b) = sqrt(a*b) does not apply to cases where a, b < 0? PureRumble 23:29, 22 November 2006 (UTC)

Precisely. You can't avoid this problem by choosing

{\sqrt {-1}}=-i

instead. "Defining"

{\sqrt {-1}}

to have two values at once is IMHO not a solution.--gwaihir 23:55, 22 November 2006 (UTC)[reply]

(after edit conflict)
Square root is not a function in complex domain, but rather a multifunction — it gives two different outputs for each parameter value (except for zero):

'i² = -1' does not imply 'i = sqrt(-1)'.

Note that

'i² = -1' implies also '(-i)² = -1'

and so

'sqrt(-1) = either (i) or (-i)',

or in other words:

'i = either sqrt(-1) or -sqrt(-1)'.

Consequently the substitution "i = sqrt(-1)" is not always true.
In this case you should use both output values, one for each 'i':

-1 = i² = i×i = sqrt(-1)×(-sqrt(-1)) = -sqrt(-1 × -1) = -sqrt(1) = -1

:-) CiaPan 23:35, 22 November 2006 (UTC)[reply]

(I took the liberty of editing the original post using math formatting) As above,