Lunchtime Contest Problem Math: Round 1

The 2008 ARML questions are are now on the web. This is how it works: you get two problems at once and a total of ten minutes to work on them. The optimal strategy is to work the easier question first, to make sure you get at least one. Remember, this is hard stuff; the average score on the top ten teams was 5.3 out of 8; getting a perfect 8 is about as hard as getting a 100% Full Combo drumming "The Perfect Drug" on expert. We'll start with Individual Round question I-1 and I-2. No peeking on future questions! You have ten minutes. Begin.

Nicholas's solutions on the flip (sorry, RSS readers) ...

... Pencils down.

I-1: Let's rewrite the measurements of the angles as 50, 50+a, 50+2a, and 50+3a. Now, extend the line segment connecting angle 2 and angle 4 down to line j. We now have a triangle at the bottom of the drawing, with angles 3, 180-4, and some other non-zero value. We know that the sum of the angles ads up to 180, and that angles 3 and 4 add up to something less than 180. Now, rewriting the angle measurements we have 50+2a + (180 - (50-3a)) < 180, which simplifies to 180 - a < 180, which further simplifies to a > 0. So, this doesn't tell us anything new.

But, the problem states than angle 3 is less than 90 degrees. Rewriting this as 50 + 2a < 90, we can solve for a, which leads to a < 20. Now, what is the maximum integer value for angle 4, which is 50 + 3a, given that a < 20? Since 20 = 60/3, we can set a to 59/3, leading to a maximum value of 50 + 3*(59/3) = 109. (The four angles, if you're curious, are 50, 69 and 2/3, 89 and 1/3, and 109).

I-2: This is the easier question which I did first. Once you write the prime factorization for a number, you can easily compute the number of factors. Take 72, which is 2³3². Add one to each exponent, and then multiply those numbers together, and you'll get the number of factors, (3+1)(2+1) = 12. You can check this: the factors are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, and 72. So, to have four factors, a number must be of the form p³ or pq for primes p and q. From here, hopefully we can solve it by using brute force. We also know that at least one of the three consecutive numbers has to be even, so it has to be of the form 2p for some odd prime p. This narrows down the hunt: just multiply every odd prime by 2 and look at neighboring numbers:

• 6 (2*3): well, both 5 and 7 are primes, so it's out.
• 10 (2*5): nope: 11 is prime, and 9 has only three factors (1, 3, and 9)
• 14 (2*7): hmmm, 14 and 15 (3*5) both work, but neither 16 or 13 do the trick
• 22 (2*11): 21-22 is a nice pair, but 23 is prime and 20 has too many factors.
• 26 (2*13): 26 works and 27 = 9³, but 25 has just three factors while 28 has too many.
• 34 (2*17): Hmmm ... 35 is 3*5, but 36 has too many factors, but 33 = 3*11, so 33-34-35 works! They asked for the smallest number, so the solution is 33.

The first time I worked this, I got the first problem wrong, because I read the question too quickly, and assumed that all the terms in the sequence were integers. This gives you a = 19 and angle 4 is equal to 107.

Admit it, folks, you would rather watch this on espn2 than a bunch of twelve-year-olds spell out-of-use words and medical terms.

Comments

[low-tech cyclist]

What happened to the m multiple in I-1?

[Nicholas Beaudrot]

Uh, m is notation for "measurement", when you see it prior to an angle.

[low-tech cyclist]

Your answer to I-2 is wrong. Once you treat 1 as a factor of an integer, you can regard as many multiples of 1 as you like as a factor of that integer, so no positive integers are quadly, since no positive integer has exactly four factors of 1; they all can have more. (Nowhere does the wording of the problem specify distinct factors, and if it did, that would have created other problems of interpretation.)

I'd have regarded 33, 34, and 35 as each having two factors, which made it a damned sight harder to work the problem.

[low-tech cyclist]

Uh, m is notation for "measurement", when you see it prior to an angle.

I'm not sure I've ever seen that notation, anywhere from elementary school through grad school, or when I taught those 'math for teachers' courses.

[Nicholas Beaudrot]

There's only one 1.

Contestants are given a series of clarifications on stuff like this the night before. I think factors are one of the things covered.

[low-tech cyclist]

I hope the kids who take this are coached well on notation and definitions that the ARML uses. Because if you can't tell what the problem is, you can't solve it.

[Nicholas Beaudrot]

I believe the term you are looking for is something like "proper factors" or something, which excludes 1 and the number itself.

[low-tech cyclist]

Simulpost!

[low-tech cyclist]

I believe the term you are looking for is something like "proper factors" or something, which excludes 1 and the number itself.

You're right. It's just that after a certain point, nobody bothers with the 'proper' anymore, because those are the only factors you're interested in anyway.

For perfect numbers, one includes the 1 (but obviously not the number itself) as a factor, but that's about the only situation I can think of where I've seen the 'improper' factors treated as factors in a few eons.

[Sir Charles]

You guys are scaring me.

[Nicholas Beaudrot]

The formula we're all taught is "add one to each of the exponents and multiply", which of course implies that 1 and the number itself count.

[Matt Weiner]

In I-2, you only have to look at 2p + 1 and 2p - 1 for all primes p; 2p + 2 and 2p - 2 are 2 (p + 1) and 2(p -1) respectively, and since p is odd p + 1 and p - 1 are even and so not prime (given that you ruled out 4 real quick).

[Sachin Gulaya]

"How many positive integers less than 100 have exactly 4 odd factors and no even factors?"

Ok, I started thinking about the problem if you add the condition that it has to be distinct factors and you allow 1 to be a factor.That problem could then be restated as follows: How many integers less than 100 can be expressed as the product of two odd prime numbers. Since the number itself and 1 are ALWAYS going to be factors and we can't have more than two prime factors then we just need the other two factors to be prime.

I started thinking about how you can count the factors of a number. First you get the prime factorization. Then the total number of factors is equal to the exponents of the prime numbers+1 all multiplied by eachother. So for 125 = 5^3 it has (3+1) factors: 1, 5, 25, 125. For 72 you have 2^3 * 3^2 so you have exactly (3+1)*(2+1) factors giving you 12 factors in total: 1, 2, 3, 4, 6, 8, 9, 12, 24, 27, 36, 72,

This is essentially counting the combinations of different factors you could make from its prime factorization. You add 1 because you need to count the fact that you can raise any exponent to the power of 0.

So now we know that we are looking for numbers from 1 to 99 that are the product of two distinct prime numbers that are ODD. So we can eliminate 2 even though it is prime(pardon the pun). 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, and we stop there because the next prime is 37 and 3*37 > 100.

So now our question is how many unique pairs of prime numbers can we pick from this list that when multiplied are less than 100. Here is where it becomes a turn the crank problem. NINE can be made with the number 3, FIVE can be made with the number 5(excluding 3*5 since we already counted that with 3), and TWO can be made with the number 7. So our result is 9+5+2 = 16 numbers.

15, 21, 33, 39, 51, 57, 69, 87, 93
35, 55, 65, 85, 95
77, 91

So yay all of our answers so far are unique.

But we forgot one thing. What about prime*(prime^2)?

Now it's more of a "turn the crank" problem than before.

3*(3^2) = 27. The factors are 1, 3, 9, 27

That's the one we left out. We can't do prime*another_prime^2 because then we have more than 4 factors.

I really don't like this method. Did anyone do it a better way?