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Course: Math for fun and glory > Unit 4
Lesson 1: 2003 AIME- 2003 AIME II problem 1
- 2003 AIME II problem 3
- Sum of factors of 27000
- Sum of factors 2
- 2003 AIME II problem 4 (part 1)
- 2003 AIME II problem 4 (part 2)
- 2003 AIME II problem 5
- 2003 AIME II problem 5 minor correction
- Area circumradius formula proof
- 2003 AIME II problem 6
- 2003 AIME II problem 7
- 2003 AIME II problem 8
- Sum of polynomial roots (proof)
- Sum of squares of polynomial roots
- 2003 AIME II problem 9
- 2003 AIME II problem 10
- Trig challenge problem: area of a triangle
- 2003 AIME II problem 12
- 2003 AIME II problem 13
- Trig challenge problem: area of a hexagon
- 2003 AIME II problem 15 (part 1)
- 2003 AIME II problem 15 (part 2)
- 2003 AIME II problem 15 (part 3)
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Sum of factors of 27000
Created by Sal Khan.
Want to join the conversation?
- How do we know there are no repeated positive integers in your matrix? (so we won't be counting a divisor more than once)(25 votes)
- Fundamental theory of arithmetic states that each number has a unique prime factorization. Let's say that a=2, b=3, c=5. Then we are looking for every unique combination of a,b, and c. So a, b, c, ab, ac, bc, abc, a2b, ab2, a2c ac2, etc... if the grid has on one row 1, a, a2, a3 and the other row 1, b, b2, b3 then were going to get every combination of a and b and we can see that were going to get a grid of unique factorizations and therefore unique numbers. Let's just do this by example for the non-believers.
1 b b2 b3
________________
1 |1 b b2 b3
a |a ab ab2 ab3
a2 |a2 a2b a2b2 a2b3
a3 |a3 a3b a3b2 a3b2
SO these are all of the factorizations with just combinations of a and b. All of these factorizations are unique factorizations and therefore unique numbers.
If we can agree that all the numbers on the first grid plane are unique then we know that all the numbers multiplied by another prime factor c will also be unique as will the ones multiplied by c2 and c3 since these are all unique prime factorizations.(40 votes)
- What does a 3 with a little 3 on it mean?(3 votes)
- it means that three has three extra brothers whose name is also 3. and multiply the three brothers(5 votes)
- Doesn't the solution provided skip the factors of 27000 that are made by combining the 5's and the 2's only as well as those that combine the 5's and the 3's only?(2 votes)
- They are not skipped. For instance if there are no 3's then that is accounted for by 3^0=1, which corresponds to the second coloumn of the matrix.(4 votes)
- how do you know what order you putt the steps in(2 votes)
- What if there was a remainder? For example, how many positive integers would there be that leaves a remainder of 4 when divided by 100? How would you solve that?(1 vote)
- There are an infinite number. Consider any number M such that M = 100*n plus 4. When divided by 100, this number will have a remainder of 4. There is no limit to n, so there is no limit to the number of M's which would have such a remainder.(3 votes)
- This is a great video by the way. It really has helped me a lot! I was just wondering why you use only 2&3 for the table but not 5. Thank you.(1 vote)
- What happens to the factors with no 2 or 3 in them?(1 vote)
- That would correspond to the first row, first column of the matrix: i.e: 1X1. This has no factors of 2 or 3 in it when multiplied by 5(1 vote)
- I've got an easier way: check this website!
https://math.stackexchange.com/questions/163245/finding-sum-of-factors-of-a-number-using-prime-factorization(1 vote) - Why would you say that there are zero 5's, when 27,000 is factored into 2's 3's AND 5's?(1 vote)
- why do i think this is too hard even for high school students?(1 vote)
Video transcript
Calculate the sum of all
positive divisors of 27,000. The easiest thing that
I can think of doing is first take the prime
factorization of 27,000, and then that will help us
kind of structure our thought of what all of the
different divisors of 27,000 would have to look like. So 27,000 is the same thing
as 27 times 1,000, which is the same thing as 3 to the
third times 10 to the third, and 10 is, of course, the
same thing as 2 times 5. So this is the same thing
as 2 times 5 to the third, or it's the same thing as 2 to
the third times 5 to the third. So 27,000 is equal to
2 to the third times 3 to the third times
5 to the third. So any divisor of 27,000 is
going to have to be made up of the product of up to three
2's, up to three 3's, and up to three 5's. So let's try to look
at all the combinations and think of a fast
way of summing them. So let's just say it
has no fives in it. It has no fives in a divisor. So if it has no fives, then
it could have up to three 2's, so let's say
it has zero 2's. So I'm just going to
take the powers of 2, so if it has zero 2's,
then we'll put a 1 here, if it has two 2's, it
has to be divisible by 4. If it has three 2's, it's
going to be divisible by 8. When I say three 2's, I
mean 2 times 2 times 2. Now, let's do it with the 3's. If you have, oh wait,
I forgot a power. If you have zero 2's, that means
it's just divisible by 1 from looking at the 2's. If you have one 2, it's
divisible by just 2. If you have two 2's,
you're divisible by 4. And if you have
three 2's, and when I mean that I'm saying
2 times 2 times 2, that means you're
divisible by 8. Let's do the same thing with 3. From the point of view of the 3,
if you have no 3's, that means at least you're divisible by 1. If you have one 3, that
means you're divisible by 3. Two 3's, or 3 times 3 means
you're divisible by 9. If you have three 3's, it
means you're divisible by 27. So let's look at all of
the possible combinations. And for this grid that I'm
going to generate right here, we assume that you're
not divisible by 5, or you're only divisible
by 5 to the zero power. So what are all the
possible numbers here? Well, you have 1 times 1 is 1. That's divisible by 1 and 1. You have 1 times 3, which
is 3, 1 times 9 which is 9, 1 times 27 which is 27. So these are all
the numbers that are divisible by that have up
to three 3's in them, from zero to three 3's in them, and
they have no twos in them. If you throw
another two in here, you're essentially going to
multiply all of these numbers by two. If you throw
another two in here, you're going to multiply
all of these numbers by 2. Now, before I do
this, because I want to do this as fast as possible. I could figure out
what these numbers are, I could multiply them. But instead, let's
just take the sum. Let's just take the
sum here of this row, of this first row
that we just did. We have 1 plus 3 plus 9
plus 27, 3 plus 27 is 30, 1 plus 9 is 10. So this is going to be 40. Now, whatever these numbers are,
they're all going to be 2 times these numbers. So the sum is going to be
80, and the sum over here is going to be 2 times
the previous row. Because here we multiplied by
2, here we're multiplying by 4, so it's going to be 160. And over here, we just
multiplied by 2 again, it's going to be 320. Or another way of thinking about
it, whatever the sum is here, it's going to be eight times
the sum of the first row. And I could, just so
you know what I'm doing, I could actually
put numbers here. This number would be 8, 24,
72, and whatever 8 times 27, I was at 160, 160
plus 56, so it's 216. But we don't want to do that. We just have to
think about the sums. So if you think about all of the
dividers of 27,000 that are not divisible by 5-- so
they're only divisible by 5 to the zero power, I
guess you could say it. We've now figured
out their sum, it's going to be the sum
of all of these rows. So if you take 40
plus 80, you have 120 plus 160 is 280
plus 320 is 600. So this is the situation. This is the sum of all of
the combinations of the 2's and the 3's that don't
have any 5's in them. Now, if you took
the same combination of 2's and 3's, so
these added up to 600, let me write it over here. So no fives Now, if you
did the same exact thing that we just did
here, but we just multiplied everything by 5. So we'd then be looking
at all the combinations that have this many twos and
this many threes, and one five, what would happen to this sum? Well, we would just
multiply it by 5. So let's multiply that by 5. So you multiply 600 by 5,
you get 30 with two zeroes, and so this is one 5 in
the prime factorization of the divisors. Now, if I wanted two 5's, I
could just multiply by 5 again. So if I multiply by 5
again, I get 15,000. This is two 5's. Another way of
thinking about this, if I just multiply
every term here by 25, which is essentially
multiplying by 5 times 5, this sum is going to be 600
times 25, which is 15,000. Now, if I have three
5's than I could just multiply this by five again. 5 times 15 is 50 plus 25 is 75. So its 75,000. So now I know all the sums. If I have no 5's, the
sum of all the divisors is 600, if I have one 5,
3,000, so on and so forth. If I want the sum of
everything, I just take the sum of these numbers. Let me scroll down a little bit. So I get, well, I
have zero, zero, than in the hundreds
place, I only have a six, and then 3 plus 5 plus 5 is 13. Is that right? Yeah, that's 13. And then carry the one,
and then I have a 9. So 93,600. So the sum of all positive
divisors of 27,000, 93,600. Hope you found
that entertaining.