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We're on problem 207. If n is equal to 4p, where p is
a prime number greater than 2, how many different positive,
even divisors does n have including n? So how many integers go
into n, or go into 4p? Well, definitely 4 goes into it,
because this is going to be a multiple of 4. Then anything that goes into
4 will also go into it. So 2 goes into it. 2 is also a divisor. p would go into it. Are there any other numbers
that go into p? No, p is prime. Well, you know 1 goes into p. But they wanted to know
the different positive, even divisors. So, the positive-- we're just
dealing in the positive world. That just lets us know this
isn't some trick question where we have to count
negative numbers. But they want even divisors. So, first of all,
1 is not even. So that can't be it. The second question,
is p even? Well, by definition, if p was
even, and if it's greater than 2, it would be divisible by 2. So it wouldn't be able
to be prime. But they tell us that
p is prime. P is prime and it's
greater than 2. So it can't be divisible by 2. So it has to be odd. So this is also not
an even divisor. So there's only two even
divisors, 4 and 2. So the answer is 2. A. Problem 208. S is a set containing
9 different numbers. T is a set containing 8 numbers,
all of which are members of S. Which of the following
statements cannot be true? The mean of S is equal
to the mean of T. So, let's think about
this a little bit. So, choice A, that their
averages are equal to each other. Let's just say that set S is
equal to 1 through 9, 1, 2, 3, 4, 5, 6, 7, 8, 9. So what would its mean be? Well its mean, if you average
all these numbers, if you average 1 and 9, you get 5. If you average 2 and
8, you get 5. If you average 3 and
7, you get 5. You average 4 and
6, you get 5. If you average 5, you get 5. So the average is 5. So if this is S, what
if T were this? What if T were just
this right here? Let me label that. That's T. That's not 7. That's T. Then S is everything. So, it's T plus the 5. So the average of all the
numbers in T, if you average 1, 2, 3, 4, 6, 7, 8, 9, your
average is going to be 5. If you think about it, the
average of 4 and 6 is 5. The average of 3 and 7 is 5. I think you get the idea. You could work it
out if you like. If you were to take set S, which
is all of these numbers plus the number 5, if you add
5 to a set of numbers whose average is 5, the average
is still going to be 5. So A definitely works. B, The median of S is equal
to the median of T. Well, once again, that's
true with this set. This is a good set
to work with. It proves everything
we need to prove. What's the median of T. It's the middle number in T. There's 8 numbers, so
you have to average the middle two numbers. So, if you average the middle
two numbers in T, 4 and 6, you get 5. So 5 is median of T. Then what's the median of S? Well, now S includes the 5. So you can put all these
numbers in order. So 5 is the middle number. So 5 is also the median of S. So, both of these can
happen, A or B. Choice C, the range of S is
equal to the range of T. The range of a set is just the
difference between the highest and lowest number. Then the range of S
is equal to what? The highest number in S is 9. The lowest number in S is 1. So that equals 9 minus 1,
which is equal to 8. What's the range of T? Well the highest number
in T is 9. The lowest number is 1. 9 minus 1 is equal to 8. So this one can definitely
be true. Statement D, the mean of S is
greater than the mean of T. Well, sure. Instead of this being a 5,
what if this was a 50? Now all of a sudden the average
of T would still be 5. But if you threw this 50 in
there for S, your average is going to go a good
bit above 5. So statement D could definitely
work if you just make that extra number that's
in S a lot larger number. Then statement E, which I'm
guessing is the choice it cannot be true because we could
figure out a case for all of A through D. The range of S is less
than the range of T. So the range of the difference
between the highest and the lowest. No matter what, the
range of S has to be at least as large as the range of T. Think about it. In this case, the range of T
is 9 minus 1, which is 8. You could make the range of
S larger if you use 50. If 50 was the extra number, then
the range is 50 minus 1. But considering that all of
these numbers in T are also in S, the difference between the
highest number in T and the lowest number in T has to be
less than or equal to the range of S. So they're saying the opposite,
the range of S is less than the range of T. We know this cannot be true
because T is a subset of S. So the choice is E. Next problem-- it took me more
time that I wanted to-- 209, How many different positive
integers are factors of 441? Well, essentially it's
just factoring it. Well let's see, this should be
divisible by-- let's see, 4 plus 4 is 8 plus 1 is 9-- so
this should be divisible by 3. So 3 goes into 441. 1 times 3 is 1. Bring down a 4, it goes
in it 4 times, 12, and then you get 21, 27. So, you get 3 times 147. 3 should also go into this,
because 1 plus 4 plus 7 is 12, which is divisible by 3. So 3 goes into 147. 147, you go 4 times
12, 27, 49 times. So let's just do a prime
factorization of 441. You get 3 and 147. 147 factors into 3 and 27, which
then factors into 3 and 9, which then factors
into 3 and 3. OK. So I didn't realize, but
this is a [? link ?]. This is the same thing
as 3 to the fifth. Is that right? Oh no, sorry. I made a mistake. Let me redo this. 3 goes into 147 not 27
times, but 49 times. So then this is, if
you you prime factorization, a 7 and a 7. I was about to say, 441 isn't
3 to the fifth power. So let's write that down here. Let me do it in a
different color. I don't want to be confusing. So 441 is equal to 3 times
3 times 7 times 7. So if they want to know how
many different positive integers are factors, so all you
have to think about is how many positive integers
can I construct with two 3's and two 7's? So let's just list them. 3 definitely works. 3 times 3 will definitely
work. Well, let's go in order. 7 will definitely work. 3 times 3 will work. That's 9. 3 times 7 would work. 21. 3 times 49 will work. 3 times 7 times 7, so 3 times
49 will work, which was actually 147. 147 works. Then 9 times 7 would
work, which is 63. 9 times 7. Then you have 9 times
7 times 7. That's essentially 441. Let me make sure I haven't
missed any combinations. You have one 3 and one 7. So you have a 3 and a 7. You have a 3 and two 7's. You have two 3's and a 7. Then you could have two
3's and two 7's. Then, of course, you have
the 3 and the 7. So I have 1, 2, 3,
4, 5, 6 numbers. Here I have 1, 2, 3-- what
am I missing here? So, I have a 3, 7--
oh, of course. I'm actually missing the 3 times
the 3, and I'm missing the 7 times the 7. So, it's 1, 2, 3, 4, 5, 6, 7,
8 factors of which 441-- and that's not one of the choices. How many different positives
integers? Oh, and of course,
the number 1. So, how many does
that come to? 1, 2, 3, 4, 5, 6,
7, 8, 9 numbers. So, that is D. There must be a faster
way of doing that. I'll let you think
about that one. Anyway. See you in the next video.