Little League Probability

I think I say it every time I post brainteaser answers, but let me again apologize for taking so long to do so.

I got correct answers from Dave, Robyn, Denise, and Giovanni, who was kind enough to point out that the brainteaser is simply an example of a binomial probability distribution.

Thanks to all who submitted answers. Here was the brainteaser:

Your favorite little league baseball team is playing for the championship. The team will play seven games (all seven games) against an opposing team. Assuming ties are not allowed, and the result of each game is dictated purely by chance, what is the probability that your team will win exactly 4 of the 7 games?

I closed my eyes, pointed my finger at the computer screen, and randomly chose Robyn's correct answer:

Think of each possible series outcome as a list of seven W's or L's.

There are 2⁷, or 128, possible arrangements since there are 2 choices (W/L) for each of 7 slots (games).

To compute how many of those contain exactly 4 W's, we need to count how many different sets of 4 slots there are that could be the locations of the 4 W's. We compute that a 7C4 or 7!/(3!4!) = 35.

So the probability of winning exactly 4 of 7 games is 35 / 128, or approximately 27.34%.

The next brainteaser is up. I said before that I wanted to do a series of probability brainteasers, but unfortunately I misplaced them. Instead, you'll get—well, just go check it out.

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Three 3's

We got a boatload of responses to the most recent brainteaser—what we are calling the "three 3's" brainteaser. Here it is:

Write three different mathematical expressions that have the same value as 9. The hitch? Or, should I say, the hitches? (1) You must use exactly three 3's and no other digits. (2) You can't use plus signs.

I'm just going to list all of the responses. Naturally, you will see some repeats. Interestingly, two different people sent in the incorrect response sqrt(3³) ÷ sqrt(3). I assumed they meant to write a multiplication sign instead of a division sign or fraction bar, so I just made the adjustment in their answers.

Also, for those of you who submitted more than three, note that I just chose three expressions randomly from those you submitted. If you don't see an expression here that you're particularly proud of, please leave it in the comments. Well done, all!

Sailorman: 3 &minus (-3) &minus (-3); (3³) ÷ 3; |(-3) &minus 3 &minus 3|

Jonathan: (3³) ÷ 3; sqr(3 × 3³); 3! &minus (3 &minus 3!)

Matthew: (3! &minus 3) × 3; 3^{(3! ÷ 3)}; sqrt(3) × sqrt(3) × 3

Denise: (3³) ÷ 3; 3^{(3! ÷ 3)}; 3 × sqrt(3 × 3)

Vlorbik: (3³) ÷ 3; 3 × sqrt(3 × 3); max{3, 3 × 3}

Dave: 3^{(3! ÷ 3)}; (3³) × (.3 . . .); 3 ÷ (.33 . . .)

Maria Miller: 3 × sqrt(3) × sqrt(3); 3³ ÷ 3; sqrt(3)³ × sqrt(3)

Mathmom: 3 &minus (-3 &minus 3); 3³ ÷ 3; 3^{(3! ÷ 3)}

Giovanni: (3³) ÷ 3; 3 × (3! &minus 3); 3^{(3! ÷ 3)}

Bogusia: 3³ ÷ 3; (3! &minus 3) × 3; Sigma (from i=3 to 3) 3i

Mike: (3³) ÷ 3; 3 ÷ .33 . . .; Sigma (from j=3 to 3) 3j

The new brainteaser is up. This one seemed to me to be a somewhat interesting probability problem. Enjoy!

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Soccer Tournament Brainteaser

We’ve got two fine answers to the most recent “soccer tournament” brainteaser. I’ve posted them below. But first, here’s the brainteaser:

Once there was a soccer tournament with only three teams. Each of the teams played one time against each of the other two. Each team scored only one goal in the entire tournament. The winner of a match got 2 points, the loser got 0 points, and 1 point was given to each team in a tie.

The results were as follows: Cardinals (3 points), Cougars (2 points), Orioles (1 point). What was the score in each match?

Stephen really went above and beyond in his explanation. I had to shorten it a bit for the purposes of this post, so I hope I captured the essence of it.

Note from the list that each team plays two matches. The winner of a match gets 2 points. A team getting a tie gets 1 point. A loser gets no points at all. Each team scores exactly one point over their two games.

Since the Cardinals got 3 points, one of the two matches must have been a win. The other match must have been a tie. Since the Cougars got 2 points, they could either have won one and lost the other, or they could have tied both matches. Since the Orioles got 1 point, they must have tied one and lost one.

Since the Cardinals won one and tied the other, some other team must lose one and tie one. The Orioles must do one of each, taking one of them from the Cardinals. Imagine all the possibilities for the two Cardinals vs. Cougar matches:

(a) Cardinals 0, Cougars 0: Cardinals tie, (b) Cardinals 0, Cougars 1: Cardinals lose (can't happen), (c) Cardinals 1, Cougars 0: Cardinals win, (d) Cardinals 1, Cougars 1: Cardinals tie.

Choice (b) can be eliminated because the Cardinals didn't lose. Given those possibilities, what are the choices for Cardinals vs. Orioles?

(a) Cardinals 0, Cougars 0: Cardinals tie, or (a) Cardinals 1, Orioles 0: Cardinals win. Nothing else makes sense for (a), because the Cardinals have to score a point in one game, and also must win the second game given the first was a tie.

(c) Cardinals 1, Cougars 0: Cardinals win, or (c) Cardinals 0, Orioles 0: Cardinals tie. The Cardinals must score 0, given that they scored 1 against the Cougars. They must tie the Orioles, so the Orioles must also score 0.

(d) Cardinals 1, Cougars 1: Cardinals tie, or (d) Cardinals 0, Orioles ?: Cardinals ? The (d) possibility can be eliminated because if they tie the Cougars, they must win against the Orioles, but if they scored a point against the Cougars, they can't score against the Orioles - but they can't win against the Orioles if they don't score a point against them.

So, let's see what happens if we consider the third match. Start with the (a) first:

(a) Cardinals 0, Cougars 0: Cardinals tie; (a) Cardinals 1, Orioles 0: Cardinals win; (a) Cougars 1, Orioles 1: Cougars tie. The Cougars had to get 1, because they got 0 in their first match. The Orioles had to get 1, because they got 0 in their first match. There are no other choices.

Here, the Cardinals get a win (2) and a tie (1) = 3. The Cougars get a tie (1) and a tie (1) = 2. The Orioles get a loss (0) and a tie (1) = 1. That sounds like the answer. But let's just check the other possibility:

(c) Cardinals 1, Cougars 0: Cardinals win; (c) Cardinals 0, Orioles 0: Cardinals tie; (c) Cougars 1, Orioles 1: Cougars tie.

Here, the Cardinals get a win (2) and a tie (1) = 3. The Cougars get a loss (0) and a tie (1) = 1. Since
they actually got 2, this eliminates (c) as a possibility.

Therefore, the (a) solution is the only correct solution: Cardinals 0, Cougars 0; Cardinals 1, Orioles 0; Cougars 1, Orioles 1.

MathNotations is a new edition to the blogroll here. It seems to me that it is truly a jewel of a site for mathematics educators and anyone interested in mathematics education. (A great set of commenters there, too—which, I think, makes or breaks a site.) Anyhoo, the author of MathNotations, Dave, is, among other things, a wonderful moderator of discussion. He offers the solution below:

For the purposes of this explanation, I will label the team as follows: A: Cardinals, B: Cougars, C: Orioles. From the description of the tournament, there were three matches: A played B, B played C, C played A.

Since A scored 3 points, it must have had 1 win and 1 tie. Since B scored 2 points, it could have had 1 win or 2 ties. Since C scored 1 point, it had exactly 1 tie.

I'll start with C because it had one tie, one loss and scored one goal. Their tie game had to have ended 1-1 (if it ended 0-0, then in it's other game it would have had to score one goal and the winning team would have to score 2, contradiction). If its tie game was against A, then A would have scored its only goal and it could not have had a win. Therefore, C tied B and the score was 1-1. It also follows that C lost to A by a score of 1-0. Finally, A must have tied B by a score of 0-0.

Summary: (1) A played B. Result: Tie. Score: 0-0. (2) A played C. Result: A won. Score: 1-0. (3) B played C. Result: Tie. Score: 1-1.

The next brainteaser is up. This one is, again, not REALLY about solving problems in mathematics. Hey! It's summer!

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Congruent Squares Brainteaser

Rory found the correct solution to this week's brainteaser using vector math and coordinates. And Jonathan used side lengths.

But I'll do something different this week and publish the solution that I have for the brainteaser. Parts of this solution are similar to what Jonathan did in finding the correct answer. (Correction: The solution provided below is pretty much exactly what Jonathan did.)

The two congruent squares shown in the image below share point D as a vertex. What is the measure of angle ACF?

Since line segments DC and DG are congruent, angles DCG and DGC must be congruent. And since the measure of angle CDG 60 degrees, triangle DCG is an equilateral triangle, and all of its angles measure 60 degrees (2x + 60 = 180; x = 60). This, of course, also means that all of its side lengths are congruent.

The measure of angle CGF, then, is 60 degrees + 90 degrees, or 150 degrees. And, since line segments CG and GF are congruent (because we found that CG is congruent to all the square side lengths), this makes triangle CGF an isosceles triangle. That means the measure of angles CFG and FCG are congruent and equal to 15 degrees each.

Since the measure of angle FCG is 15 degrees, and the measure of angle DCG is 60 degrees, then the measure of angle DCF must be 45 degrees. And since the measure of angle ACD is 45 degrees, the measure of angle ACF is 45 degrees + 45 degrees, or 90 degrees.

I'll post Rory's and Jonathan's solutions in the comments. The next brainteaser is up.

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More Than One Way to Skin a Brainteaser

I’ve got a pretty nice geometry brainteaser for you this week, but let me first finally share with you some correct answers to the most recent brainteaser.

Visualize:
A: 1 .... 7 .... 21 .... 43 ... a. B: 2 ... 10 ... 26 .... 50 ... b.
C: 3 ... 13 ... 31 .... 57 ... c. D: 5 ... 17 ... 37 .... 65 ... d.
Can you find a, b, c, and d?
It’s interesting that there are so many correct ways to think about this problem. First up is just a part of Jonathan's answer.

I get the same answers going across or going down: 73, 82, 91, 101.

Down: 1, 2, 3, 5, 7, 10, 13, 17, ...
Try again: 1, +1, +1, +2, +2, +3, +3, +4, ...

Across: A. The nth member of A = 1 + 2n(2n &minus 1), so 1, 7, 21,...
Across: B. The nth member of B = 2 + 4n(n &minus 1).
Across: C. The nth member of C = 4n² &minus 2n + 1.
Across: D. The nth member of D = 4n² + 1.

And Rory, who now belongs to the 10% of the population that finds the correct answer to the Wason Selection Task, came up with a similarly elegant solution. I won't reproduce his diagrams here. His solution needs only one sentence:

The difference between each number and the one behind it increases by 8 each subsequent number.

So, for A, we have 1 and 7 with a difference of 6, then 7 and 21 with a difference of 6 + 8, or 14, then 21 and 43 with a difference of 14 + 8, or 22, etc. This applies to all the patterns—A, B, C, and D.

Here is the solution I had, and it's why I included the word visualize. My apologies for some of the interior lines not displaying:


Enjoy this week's geometrical brainteaser.

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Circular Permutations

It's getting so I needn't mention anymore who answered the brainteaser. Here they are. Surprise, surprise:

In how many ways can 5 children sit around a circular table if 2 brothers are always (a) to sit next to each other, (b) separated?

Circle permutations are fun to teach kids. There will be a formula, yes, but really, no one in the room (besides me) already knows it, and, if you learn the principles involved, you never need to actually learn the formula anyhow. Certainly memorization is not crucial.

(a) We have girls L, M, and N, and brothers C and D. First, seat C, anywhere, doesn't matter (one way to do that, just sit). Next, seat D (to C's left or right, 2 ways). Next, seat L (3 open seats). Next, seat M (2 open seats). Finally, seat N (1 open seat): 1 × 2 × 3 × 2 × 1 = 12 ways to do this.

(b) To seat the brother's separated, two methods.

First method: All the ways to seat all 5 &minus all the ways to seat the brothers together = the ways to seat the brothers separately: 1 × 4 × 3 × 2 × 1 &minus 12 = 24 &minus 12 = 12.

Second method: First, seat C, anywhere, doesn't matter (one way to do that, just sit). Next, seat D not next to him (two ways to do that). The rest as in (a), above. -- Jonathan

If there were no restrictions, there would be 24 unique ways to arrange 5 children around a circular table. There are fewer unique circular arrangements compared to linear arrangements (120) because shifting positions around a circular table results in identical
arrangements.

(a) If two brothers (a and b) must sit next to each other, there are two ways they may be seated (a, b and b, a). The other three children can be arranged 3! or 6 different ways. This gives 2 × 6, or 12, seating arrangements in which the two brothers are together.

(b) To find the number of arrangements in which the two brothers are apart, subtract the answer in part a from the total number of arrangements. This gives 12 seating arrangements in which the brothers are apart. -- Colleen

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Permutatin'

Two esteemed persons correctly answered this week's permutation brainteaser imported from Singapore. Here it is:

In how many ways can 4 boys and 2 girls seat themselves in a row if (i) the 2 girls are to sit next to each other, (ii) the 2 girls are not to sit next to each other, (iii) the 2 girls are to be separated by 2 boys between them?

Here's Jonathan:

Boys: pqrs, Girls: yz

(i) For 2 girls together, rearrange pqrs(yz) or pqrs(zy), 5! + 5! is 240 ways. Alternate, seat one girl. Case I, she's on an end. Then there is no choice where the second girl sits, and 4 × 3 × 2 × 1 ways to seat the boys. Case II she's in the middle. Then there's 2 places for the second girl to sit, followed by 4 × 3 × 2 × 1 ways to seat the boys. So we have 2 × 1 × 4 × 3 × 2 × 1 + 4 × 2 × 4 × 3 × 2 × 1 = 48 + 192, = 240 ways.

(ii) For girls separate, take the total arrangements (6!) and subtract the previous answer 720 &minus 240 = 480. Alternate (same cases as above): 2 × 4 × 4 × 3 × 2 × 1 + 4 × 3 × 4 × 3 × 2 × 1 = 192 + 288 = 480.

(iii) For girls separated by two boys, let's seat one girl first (6 ways), then the second girl (uniquely determined.... seat 1 --> 4, 2 --> 5, 3 --> 6, 4 --> 1, 5 --> 2, 6 --> 3) , then the four boys (4 × 3 × 2 × 1). 144 ways.

And here's Colleen:

(i) Since the girls must remain together, they occupy a
single position in the row. Together with the boys, there are 5 positions that must be arranged. The number of arrangements would be 5! or 120. Within each of these arrangements, the girls' positions can be reversed giving a total of 240 unique arrangements.

(ii) If there were no restrictions at all, there would be 6! or 720 unique arrangements of the six children. If we subtract from that the number of arrangements in which the girls are next to each other, we would be left with the arrangements in which they are apart. 720 &minus 240 (from answer 1) = 480 unique arrangements.

(iii) The only positions the girls can occupy are 1 and 4, 2 and 5, or 3 and 6. In each case, the girls can be seated two different ways for a total of 6 arrangements. The boys can be arranged 4! or 24 different ways. Therefore, there must be 6 × 24 or 144 unique arrangements.

Nice job, folks. The next brainteaser is up. One more week of permutations.

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Card Order Brainteaser

I've been doing a bad job of updating the brainteasers because of an upcoming gig in Atlanta and the imminent birth of my third child.

I'll start it up again in the end of March. Until then, enjoy these correct answers to the latest brainteaser:

There are three playing cards facedown in a row in front of you: a queen, an ace, and a six. One is a heart, one a diamond, and one a spade. The six is immediately to the right of the spade. The diamond is between the heart and the queen. Describe the order of the cards.

John at NCLBlog provides a beautifully constructed explanation:

Left to right: Queen of spades, six of diamonds, Ace of hearts.

1. From "the diamond is between the Queen and the heart," we know:

a) The Queen is the spade,
b) The diamond is in the middle.

2. Because the six is "immediately to the right of the spade," the spade can't be on the right. Since the spade also can't be in the middle (1b), it's on the left, and we know from 1a that the spade is the Queen.

3. The six, then, "immediately to the right of the spade," is in the middle and, from 1a, is a diamond.

4. That leaves the Ace of hearts on the right.

Jonathan does it in six lines:

Spade - 6 - X or X - Spade - 6 (since the 6 is to the right of the spade).

Y - Diamond - Z (since the diamond is between two other cards).

So, Spade - 6 of Diamonds - X (combining those two, spade can't be in the middle).

The Queen is not the heart (diamond is between the heart and the queen) so the heart is on the right, queen is the spade on the left.

Queen of Spades - 6 of Diamonds - X of Hearts.

(And since only the Ace is left): Queen of Spades - 6 of Diamonds - Ace of Hearts.

And finally, Giovanni solves it in words:

From left to right: queen of spades, six of diamonds, ace of hearts.

From Clue number 2: The diamond is between the heart and the queen. Therefore the queen must be spade, because it cannot be neither the diamond nor the heart. The queen cannot be in the center, because the diamond is in the center.

From Clue number 1: The six is immediately to the right of the spade. We know from the previous inference, that the spade is the queen, therefore the six is immediately to the right of the queen. The queen we knew already it was not in the center: therefore the six is in the center and is the diamond, and the queen is left of it.

Therefore the ace must be to the right, and must be a heart.

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Perfect Square Brainteaser

Here it's been nearly a month since I posted the current brainteaser, and now I'm going to show my appreciation by only posting one answer.

My apologies. I got several great responses (all correct, I might add), and I hope you continue to send in your responses in the future. When I can, I will post ALL the correct responses, so long as they have explanations.

I picked this week's winner based on the fact that it was "suggestive" of a model (I supply the models below), and, well, it made me laugh out loud. The winner is Jeremy Clark (www.site.uottawa.ca/~jclar037).

Here was the brainteaser:

There are six rows shown below. Each row has a pair of numbers and a blank line:

1, 24, ___
2, 23, ___
3, 22, ___
4, 21, ___
5, 20, ___
6, 19, ___
The numbers in each pair add up to 25, which is a perfect square. Your job is to fill in the blank lines with a third number--a different number in each row--so that the sum of ANY two numbers in the row is ALSO a perfect square.

And below is Jeremy's answer and explanation. One of the nice things about this explanation is that it builds the algebra from scratch, using a visualization.

A perfect square is easily visualized by imagining you have a bunch of pennies and you arrange them on a table in, as the name implies, a perfect square. Now depending on how many pennies you have, its possible you'll end up a few pennies short of forming that perfect square (you may be able to form a perfect rectangle; or you might find yourself always a few pennies short of any perfect rectangle--in which case the number of pennies is prime).

The first set of numbers are 1, 24, __. We'll call the first number A, the second B, and third (the one we are trying to find) C.

C will probably be a few pennies short of a perfect square. In fact, adding A pennies will make it a perfect square. So we can picture C in our heads as an almost perfect square with a hole in it--a hole that is A pennies big. For the fun of it, lets call the missing pennies C's A-hole.

The second condition is that if we add B pennies to C, we'll fill up the A-hole so it's a perfect square and then we'll add an additional L-shaped border of pennies around two sides of the square to make a new perfect square that's one penny wider and one penny longer than the filled square.

Note: The B doesn't have to be "one penny wider and one penny longer than the filled square." It could be 2 or 3 or 100. The size of B does not affect the algebra Jeremy is constructing.

The L-shape addition is made up of B pennies minus the A pennies that we used to fill the A-hole. If we drop the penny that is in the corner of the L, then we have B &minus A &minus 1 pennies. If we divide this number by two [(B &minus A &minus 1) ÷ 2], we get the length of one side of the L (without the corner) which is the length of one side of the filled square. This length times itself, [(B &minus A &minus 1) ÷ 2]^2, gives the number of pennies in the filled square. And if we remove A pennies from the filled square, [(B &minus A &minus 1) ÷ 2]^2 &minus A, we are left with C pennies. In mathematical terms:

C=[(B &minus A &minus 1) ÷ 2]^2 &minus A
So our C's are: 120, 98, 78, 60, 44, 30, (18, 8, 0, -6, -10, ...).

Nice work. I'll have the new brainteaser up this weekend. Stay tuned. And thanks again to everyone who submitted an answer.

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Train in the Tunnel

This week's brainteaser was easy enough to scare people away from answering it, I think. We did get two correct answers, though.

A train that is 100 meters long is traveling at 100 meters per minute. The train passes through a tunnel that is 100 meters long. How long does it take for the train to pass through the tunnel?

Denise from the soon-to-be-blogrolled Let's Play Math! worked past her initial fear to come up with the correct answer:

This one seems too easy, so perhaps I am missing something. Two minutes, right? It should take one minute for the locomotive to get through, but as the locomotive pulls out, the caboose has just entered the tunnel, so it takes another minute for the whole train to come back into the open.

Well done, Denise!

John from NCLBlog got the right answer as well. I think this is the first time he's sent in an answer:

Well, it takes two minutes. After one minute, the 100-meter train traveling 100 meters per minute fits exactly inside the 100-meter tunnel. That means its nose is at the front and its tail is at the back. After another minute, the 100-meter train's tail moves from the back of the tunnel to the front.

Bravo! The next brainteaser—a lot harder this time—is now up for your perusal.

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Important Numbers Brainteaser

Maria and Jonathan both got the right answer to this week's brainteaser. Yet again, Jonathan found a "creative" way to solve it.

Place these numbers (113, 355, 408, 878, 610, 987, 577, 323) into four pairs. In each pair, the higher number divided by the lower number should yield a number (to an average of five decimal places) important in mathematics.

Here's Maria's answer, short and sweet:

987/610 = 1.618032786, close to golden ratio
355/113 = 3.141592920353, close to pi
878 / 323 = 2.71826625, close to e
577/408 = 1.41421568627, close to square root of 2

I won't repeat the numbers in Jonathan's answer, but here was his method:

I cheated. I copied your numbers to Excel, used "Data to columns" to lose the commas. Paste/Special transpose gave me the border of a table, and I filled it in with a little formula, dividing each number by each other. Then I looked for famous numbers.

Nice job, folks! The next brainteaser is up.

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Morse Code Brainteaser

The most recent brainteaser asked you to make sense of some Morse code without breaks between letters or words. Alas, there were no winners.

Here is the code:

• • • • • • • – • • • – • • – • – • • – – – – – • – • • • • •
And here is the code with the appropriate spacing (between letters). The solution is HE FELT REMORSE.

• • • •   •   • • – •   •   • – • •   –   • – •   •   – –   – – –   • – •   • • •   •
Okay, back to mathematics for this week's brainteaser.

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Seeing Multiples

Dr. P. from EduInsights came up with the textbook answer to this week's brainteaser. And Jonathan got it in a completely different way.

Find the missing numbers: 31 62 ? 25 56 ? 19.
Dr. P.'s answer is succinct and methodical:

First reverse the digits: 13 26 ? 52 65 ? 91. Then note that all the known entries are the multiples of 13 in order. So the missing entries are easily supplied: 13 26 39 52 65 78 91. Restore the original sequence by reversing digits: 31 62 93 25 56 87 19. The missing numbers are 93 and 87.

Jonathan's correct answer was creative:

Looks like we are counting objects numbered from 1-99, and counting off every 31st (98, 99, 1, 2,...). So: 31 62 93 25 56 87 19.

Jonathan adds 31 in a circular way, starting over with 1 after he reaches 99. In essence, he subtracts 99 every time the sum (term + 31) is greater than 100. This progression begins to diverge from Dr. P.'s multiples-of-13 answer immediately after the last term given (19), but it is not incorrect, as it supplies the correct missing numbers just as well. A good follow-up question for the comments would be this: Why do the two answers both produce the same seven terms?

The new brainteaser is up. You have until the 5th of January. Good luck with that!

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The "New Way" to Multiply

Jonathan comes through like a champ and gives us something in response to the brainteaser, which is certainly not a "new way" to multiply.

Right, there's implied powers of ten with the lines, as with decimal representation of numbers. And each intersecting zone or nodelet or whatever we call it has a bundle of 10^m's crossing a bunch of 10ⁿ's, giving a product times 10^{(m + n)}.

Great if you want to do long multiplication without knowing the times table up to 10, but why? I don't think I would like kids spending time during this activity counting the intersections involved with 78 × 96. And this gets real ugly as soon as 0's start popping up (I know the placeholders in 'regular' multiplication are also ugly).

But just look how colorful it is and think about how hands-on it would be. You mean to tell me you wouldn't want to see multiplication done like this:


The next brainteaser is up.

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Amusia and the Brainteaser

Over at Mind Hacks, I found this really cool study, described at Cognitive Daily, regarding rhythm and amusia.

And so far, no one has sent in an explanation for this week's brainteaser to try and explain how this "new way to multiply" works, so I'll keep it up there for one more week. One of the problems with the method shown in the video is that it is done with diagonal lines, so maybe straightening things out might help to straighten things out:

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Multiples of 6

First of all, thank you all for your patience with the brainteaser these past two weeks. I am beginning to come out from under the pile before me.

Secondly, let me thank new submitters Giovanni and Mr. Peter Bonner for sending in their correct answers. I like to see new contributions.

Here was the brainteaser for the past two weeks:

Suppose n is a whole number. Explain why this expression must be divisible by 6: n(n + 1)(2n + 1).

I have to applaud Giovanni for his concision, so I'm posting his answer below. The only thing he seems to have left off (even before my edits) is a thorough answer to this question: Why, if n is 1 above a multiple of 3, must 2n + 1 be a multiple of 3? No submission really explained this to my satisfaction. But perhaps we can flesh it out in the comments here.

Divisibility by 2: either n or n + 1 must be divisible by 2.

&bull If n is even, then n is divisible by two.
&bull If n is odd, then (n + 1) must be even, thus divisible by 2.

Divisibility by 3: n can be either a multiple of three, or 1 below, or 1 above.

&bull If n is a multiple of 3, there you have the factor.
&bull If n is 1 below, then (n + 1) will be a multiple of 3.
&bull If n is 1 above, then 2n + 1 will be a multiple of 3.

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What, No Bar Models?

I'm surprised that, as of yet, no foot soldiers in the bar model army have taken up this week's brainteaser challenge. Oh well. You snooze, you lose.

A bag contains a number of balls, 24 of which are green and the remainder blue. If a ball is chosen at random, the probability that it is green is 0.8. Find the number of blue balls in the bag.

Now I suppose the challenge should be to come up with the worst possible model to explain the mathematics behind the solution. Try thinking cinemathematically. It's a steep challenge.

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Pattern Brainteaser

Colleen of Math Playground has done it again, submitting yet again the only correct answer to the brainteaser this week. Here it is:

What are the two missing numbers in the series below?
? ? 3 3 7 7 2 3 6 5

Answer: 4, 3

This problem didn't seem to fit any mathematical model so I approached it as a true brainteaser rather than a math problem. I focused on the 3,3,7,7 part of the sequence and noticed it corresponded with the number of letters in the words used in the question.

Well done. I've made the next brainteaser a lot easier. So we should be seeing some nice explanations I hope. Maybe we'll even see some nice visuals.

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Colleen's Algebra

Colleen was the only one that e-mailed me an answer to this week's brainteaser. And she got it right. The brainteaser is shown below.

A mother is 3 times as old as her daughter was when the father was the same age as the mother is now. When the daughter reaches half the age the mother is now, the son will be half as old as the father was when the mother was twice the age the daughter is now. When the father reaches twice the age the mother was when the daughter was the same age as the son is now, the daughter will be four times as old as the son is now. Given that one of their ages is a perfect square, what are the four ages?

And here's Colleen's answer, which she generously sent to me as an image:

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Running Probability

So sorry about missing the weekend deadline for posting the brainteaser winner. This week's winner is Mr. B:

Three runners compete in a race. The probability that A will win is twice the probability that B will win. The probability that B will win is twice the probability that C will win. What is the exact probability that player A will win the race?

And here's Mr. B's most excellent answer:

Let c be the probability that Runner C will win the race. Since the probability that Runner B will win (b) is twice the probability that Runner C will win, we have b = 2c.

Similarly, since the probability that Runner A will win (a) is twice that of Runner B, we have

          a = 2b
          a = 2*(2c)
          a = 4c

Now, the sum of all possible outcomes of a probability experiment is 1. Since there are only three runners, we have

          a + b + c = 1
          4c + 2c + c = 1
          7c = 1
          c = 1/7

Therefore,

          a = 4c
          a = 4*(1/7)
          a = 4/7

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