Who Wants to Be a Mathematician?

Above, contestant Alexander Zhu, host Mike Breen and winner Linus Hamilton.

About 6,000 mathematicians descended on Washington this week for a conference hosted by the American Mathematical Society. I was among them yesterday for a session on math education with some inspiring middle school teachers.

If any of you attended, I'd love to hear what you learned!

I also want to give a shout out to Linus Hamilton, a 14-year old freshman from Eleanor Roosevelt High School in Prince George's County who won first prize, $3,000, and a graphing calculator in the "Who Wants To Be a Mathematician" competition.

Overall, he said the questions were "medium hard" and that he was "really nervous." He plans to put the money into a college account.

The other contestants were:

Alexander Zhu (Poolesville High School)
Akshar Wunnava (Thomas Jefferson School of Science and Technology)
Siyun Lai (Washington-Lee High School)
Alex Golden (Springbrook High School)
Rachel Coston (Potomac Falls High School)
Rachel Zhang (Walt Whitman High School)
Michael Lindsey (St. Albans School)

Here are some sample questions from a test these students took to qualify for the contest:

How many positive integers between 1 and 101 are multiples of 5 but not multiples of 7?

Four fair coins are tossed. What is the probability that the number of heads is greater than, but not equal to, the number of tails?

By Michael Alison Chandler  |  January 7, 2009; 6:35 PM ET  | Category:  Mathletics
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Well, since no one is commenting, could you relieve the suspense and give us the answers?
How does one get the answer to the first question? You can list all the multiples of five and seven. You'll have one that is shared by both, at least: 5 times 7. Then double that and you get 2 x 5 x 7. That's two.
It seems to me that 35 and 70 are the only multiples of 7 between 1 and 101 that fit the profile of a multiple of 5.
Multiples of 5 end in a zero or a five.
Then divide 101 by 5 to get the number of multiples of 5. Subtract 2. That's your answer. 20-2 = 18.
Or I could be way off base.
I wish people would explain how they get answers. This is what is so frustrating about learning math sometimes!
For the second question, tossing a fair coin, the probablity of getting heads is 1/2. Tossing multiple fair coins doesn't change that, does it? So if you are asking what the probablity of that probablity NOT being .5, I'd say it is 0. That's assuming you just tossed 4 fair coins at once.
Even if you toss them one fair coin after the other, then since the probability of producing heads is the same as that of producing tails, there would be no difference in the two. So the probability of heads being greater would be zero.
It would be figured the same way if you were calculating the probablility for the number of tails, since tails also has the probablity of .5.
The probablity of the number of heads being greater than the number of tails would be zero. Because they would both be the same probabilty.

Posted by: KathyWi | January 13, 2009 10:35 AM | Report abuse

I'll put in my 2 cents, KathyWi.

I think your accounting for the first problem is correct and 18 is the answer. The second problem I interpreted as follows. You flip 4 coins with possible outcomes of 4H0T (4 heads and 0 tails), 3H1T(3 heads and 1 tail), 2H2T, 1H3T, 0H4T. So the outcomes that correspond to the number of heads being greater than the number of tails are 4H0T and 3H1T. The trick is figuring out the probabilities of these two occurrences. The probability of 4H and 0 tails is .5 times .5 times .5 times .5 = .0625. The .5 comes up because the probability of each of the coins coming up heads is .5

You might think the probability of 3H1T is the same, but it is not because there is more than one way to get this. you could flip HHHT, HHTH, HTHH, or THHH if you flip the 4 coins in order. Each of these four occurrences has 3H and 1Tail so the probability of each occurence is also 0.0625. But there are four such occurrences so 4*.0625 = .25. So the final probability of getting 3H1T is .25. Adding up the probabilities for 4H and 3H1T gives a final answer of .3125 or a 31.25% chance of getting more heads than tails.

Posted by: fedbert | January 15, 2009 2:35 PM | Report abuse

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