VMHS Math Circle

advertisement
VMHS Math Circle
XVI. The Grand Number Theory-ish Review (Part I: Primes and Factors/Multiples)
Number theory is the branch of math that has to deal with the set of integers. Sometimes, when
math people (excuse my terminology, there) mention number theory, it’s definitely a possibility that
they’re referring to any subset of the real numbers.
You might have heard of the theorem that took about 357 years to successfully prove: Fermat’s
Last Theorem. It stated that no three positive integers a, b, or c could satisfy the equation a n + bn = cn for
any “n” greater than two. That’s a big example of a number theory problem. Although it was so simply
stated, yet so difficult that solving it took entire centuries, the AMC will never, ever get this tough.
This number theory review will be shortened because we’ve essentially covered two big
things that are part of number theory: modular math and number bases. I wanted to put those topics
as earlier documents because those topics evade the high school curriculum. If someone just walked into
the AMC not knowing anything about Vieta’s, modular math, number bases, or mass points, he or she
will be at a disadvantage, if not downright helpless.
Make sure you know the following before we start.
 Least common multiple
 Greatest common factor
 Simplifying and adding/subtracting/multiplying/dividing fractions (there is no reason
why you shouldn’t know this already I’ve done this numerous times with practice
problems from before.
Let’s start off.
Primes
Primes are numbers whose divisors are only 1 and itself. If someone says that a negative number
“n” is prime, they most likely mean that the only divisors of that prime are 1, -1, n and -n.
Here’s a good tip for checking for primes. It involves quite a bit of long division, however.
Divide by the first few natural numbers (1, 2, 3, …). If you notice that nothing works, start dividing by
primes (3, 5, 7, …). If you notice that you don’t have integer quotients, after dividing by a lot of primes,
and that those quotients are less than the prime you’ve divided by, you probably have a prime number.
Does this sound a bit confusing? Let’s try an example.
Is the number 499 prime? If not, what are its factors?
Let’s break out some long division (I don’t know how to do that on MS word, so I’ll just hand you the
numbers.).
There is no way that the number 499 is divisible by 2 since it’s not even. That rules out all potential even
divisors. 499/3 is not an integer, so we rule out all potential divisors divisible by 3. There is no way that
499 is divisible by 5, for 499 does not end in a 5 or 0. 499/7 is not an integer. It seems like we’re not
having any luck. Let’s check all potential prime divisors, then. 11, 13, 17, and 19 are no-gos.
We try 21 and 23. Trying 21, we notice that our quotient is about 23, but is not an integer. Trying 23, we
notice that the quotient is about 21, but it’s still not an integer. Now, we’ve reached a point in which our
quotients are less than the prime we’ve divided by, upon reaching 23. We can conclude that 499 is a
prime number, since all other divisors will result in non-integer quotients that lie near the primes we
have already tried.
Before we move on, note that neither 1 nor 0 are prime or composite. However, 0 can be a
multiple of any positive number. Assume that this is true unless specified by the AMC.
Factors and Multiples
This part is pretty instinctive and self-explanatory, so there are just a few formulas to cover. Make
sure you know that the set of factors of a number “n” consists of all real numbers that, when multiplied by
another number, results in “n.” It’s pretty much easy to say that an integer “n” consists of infinitely many
real factors. However, concerning integers, which are what the AMC will look at in number theory
problems, things get much different.
Make sure you know this formula:
Total number of positive integer factors of a number “n” (including 1 and “n”):
If we take the prime factorization of “n,” and look at each exponent e1, e2, …, en of all prime factors of
“n,” the total number of positive integer factors is (Yes, I finally get to use this symbol!):
n
∏[e
k
+1
]
k=1
For example, let’s find the total number of positive integer factors for 63. The prime factorization of 63 is
32 * 7. Thus, the total number of factors is (2 + 1)(1 + 1) = 6. Checking, we see that the factors of 63 are
1, 3, 7, 9, 21, and 63. That’s six factors.
The formula above is a huge part of number theory, especially for the AIME. As for factors and
multiples, all you need to know are the meanings of GCF and LCM, as well as how you can use this to
your advantage. This will come up in some sample problems.
A positive integer “n” has 60 divisors. However, 7n has 80 divisors. What is the largest power “k” of 7k
that divides “n”?
Okay, a good thing to notice is that the prime factorization of 60 is 2 x 2 x 5 x 3, so “n” is in the form
a1a2a34a42. If we’re going to get 80, we might as well have the prime factorization 2 x 2 x 5 x 4, with the 4
there instead of the 3 since that’s the only way we can get factors that multiply to 80 by adding 1 to a
divisor of 60. What does this tell us about “n”? That’s right. From what the question gave us, its prime
factorization contains 72. Thus, our answer is 2.
Let “S” be the set of the 2005 smallest positive multiples of 4, and let “T” be the set of the 2005 smallest
positive multiples of 6. How many elements are common to “S” and “T”?
The problem tells us that S and T are both sets of positive multiples. We don’t need to worry about zero.
Carrying on, S consists of all multiples of 4 from 4 to 8020. T consists of all multiples of 6 from 6 to
12030. The least common multiple of 4 and 6 is 12, so we need to count how many multiples of 12 there
are in “S.” Why not “T”? This is because multiples of 12 in “T” that exceed 8020 are not in “S” anymore.
Finishing this off, we have ⌊8020/12⌋ = 668.
TIPS:
 When checking whether a number is prime, do it in a routine manner. By this, I mean to check the
first few positive primes. It’s obvious to rule out 2 and 5, but don’t ignore 3 or 7. If you check
with larger primes, you may end up skipping smaller numbers which involved easier
computation.
 Don’t forget that the total number of factors of an integer is what you get when you line up one
more than each exponent of each individual factor of that integer. Let’s consider our example of
finding how many factors were in 63, this time working backwards to find the prime factors given
the number of positive integer factors. We are given that the number 63 has 6 divisors, and the
prime factorization of 6 is 3 * 2. Do not make the mistake that the prime factorization of 63 is
3372! Subtract 1 from 3 and 2 to get your desired prime factorization: 327.
*Yawn* I actually ripped the last two sample problems from the actual AMC. I made it clear to myself to
create my own problems to enhance my own problem-solving strength and save the experience of
practicing with real AMC problems for you guys till later, but school’s started. Plus, I was typing this
document up at midnight and finished at 1:34 AM. I hope some slack can be cut for me. 
Download