okay this question is from the Berkeley math tournament for middle school students that's
why it's BmMT and this is from the year of 2021 and yes I'm back at UC Berkeley just for
today and maybe for a few days yeah anyway this question is actually the last question of their
individual test and the question is pretty much asking us to find all the factors for the number
of 360 and then we do one over the first Factor plus one of the second factor and so on so on so
on up to one over the last Factor you know the deal please pause the video and try this first
done cool I really like this question because it goes over a lot of the things that we should
know regarding to the factors of a number and let's just stop this question first
right here if you really want to do one over the first Factor plus one of the second Factor
no that's not a good idea because you will have to get a common denominator not a good idea right
so check this out we know that a is a factor for 360. so that means we must
have a times some other number that's going to be D together we should get 360 and this is just a
natural number and from here we can say a is just equal to 360 over D and now instead of
looking at the sum of 1 over a we can just go ahead and do what the sum of 1 over 360
over D and we can put the D on the top so this right here is just the sum of D over 360.
and this is actually still the same kind of question because when a is a factor of
360 and we know a times D is 360 this means D is almost the whole a factor of 360 isn't it
and the purpose of looking at this instead because now everybody has the same denominator namely 360.
so if we can find the sum of D then we're pretty much done we can just go ahead and divide it
by 360 afterward so of course the question is how do we find the sum of the factors of 360.
well let me show you and I will explain how we can end up with that procedure later on all
right this is what we do whenever we're talking about factors the best thing that we can do is
just go ahead and prime factor it and I'm just going to prime factor it for you guys so we'll
get 2 to the third power times 3 to the second power times 5 to the first Power because that's
8 times 9 which is 72 times 5 is 360. hmm okay well I would like to tell you guys that the sum
of all the factors so I'll just put all the factors all the factors
of 360 is the following check this out you see how we have 2 to the third power here what we
do first is just 2 to the 0th power first and then we add 2 to the first and then 2 to the second and
then 2 to the Third then we move on to the next one which is due to the second power so we start
with 3 to the zeros power plus three to the first plus three to the second and of course we do the
same thing for the last one which is 5 to the zero plus five to the first and then just go ahead and
work it out this right here will give us 1170 and in fact this right here will give us the sum of
d that's it but wow how do you end up with that right don't worry I'll explain that as
I told you guys but let's finish this question first the answer for this right here after we
have done this it's just going to be what the sum of D which is just 1170 and then divided
by the common denominator which is 360. and of course be my guess to reduce it we get 13 over 4.
and that's pretty much what the question is asking of course the official question is asking us
once we reduce the fraction just go and add them up so the final answer is 17 but
that's the main thing right here so now let's talk about
how do we find the sum of all the factors and in fact how many factors do we actually have
the okay stuff as well all right if you want to know the number of factors
of 360. we also utilize the prime factorization but this title will just focus on the powers
right here check this out what we do first is just do three plus one and then the next Factor
the next power which is two and then you add one to it and then the last one one plus one
multiply them we get 24. so we have a total of 24 factors so if you really want to put a force
solve this equation be my guess you will just have like 24 fractions and then just add them up
not so bad but anyway I don't know that's not no that's not a wonderful one here
I'm going to explain to you guys like why those procedure work with an example and I
was just kind of generalize it at the end because I have another boy after here but anyway here we
go let me just give you guys an easy example to look at so let's consider the number 40.
and then I'm just going to prime factorize it for you guys as well right here which is 2 to
the third power times 5 to the first Power and the reason I'm using 40 is because we only have
two different primes here right so two and five that's easier to look at now let's first talk
about how many factors that we have for the number 40. well let's see 40 we can actually
just write them out we know this is 1 times 40 and then we can also do 2 times 20. 4 times 10
and we can also do five times eight so we have a total of eight factors and these are it cool um
according to what we did earlier is since that we just multiply
three plus one which is four and then two and then one plus one which is two
so four times two is eight and it works well here's the deal
if you look at a factor here let's say the 20 here this is the same as what 2 to the second power
times 5 to the first Power all right and then if we look at the number eight right here this is
the same as saying 2 to the third power times 5 to the zeros power okay and if we look at the number
for here this right here is just 2 to the second power times 5 to the zeros power okay cool in fact
let's make an observation notice that all these numbers here they must be two to some power and
then also 5 to some power that's the only way right so in order for a number to be a factor
of 40 we must have two to some power let me just say 2 to the eighth power times five to some power
the factor must be in the form of this and then a has to be what either 0 1 2 was 3. right because
if you look at five here this is 2 to the 0 times 5 to the first so the power of 2 can be zero and
similarly B right here is either 0 or 1. so as we can see here we have a total of four choices
and then this right here we have a total of two choices
so if you want to find out the number of factors for 40 which is four choices times
two choices so all together we have a total of eight factors so that's
pretty much it so let me just go ahead and write this down number of factors
of 40 is just again what we did earlier right three plus one times one plus one which is just
equal to eight and then we are done cool now let's take a look at the sum of all the factors hmm
I'm just going to write them out and then add them up yeah but I'm not going to write
down one plus two plus four instead I'm going to write a 1s 2 to the 0
times 5 to the zero that's one next number is 2 which is the same as 2 to the first times pi to
the zero and then the next one is two to the two times five to the 0 and then the next one
is five which is 2 to the 0 times 5 to the first and then continue Plus 8 which is 2 to the third
times five to the 0 and then 10 is 2 to the first times 5 to the first and then 2 to the
two times five to the first and lastly 2 to the third times five to the e first as you can see
this right here should contains you guys forwarded earlier cool huh and now check this out we are
going to do more factorings but this time it's more like the algebra style have a look this
turn has 5 to the zero I know it's just equal to one but that's 2K DS 5 to the zeros and likewise
this has 5 to the zero and then likewise this and then likewise this so with that being said
we can actually factor out five to the zero and we can just put this down as 2 to the 0 plus 2
to the first plus two to the second plus two to this third and then multiply by 5 to the zero
continue we can do the same thing for the rest this right here has 5 to the first five to the
first here as well and likewise this likewise that guess what yeah this this this does not
match with this so again we can write it as two to the zero plus two to the first plus two to
the two plus two to this third times five to the first and that are the same so one second
we can Factor it so we get 2 to the zero plus two to the first plus two to the two plus two
to the third and then we multiply with this and that which is 5 to the zero plus five to the one
and of course you can work it out and notice that we only have like two sets
of parentheses multiplying because we only have two and five if we have three times to
the something power just like probability earlier then you do exactly what we did
right you see you do exactly what we did and of course you can just kind of practice
your facting skill and then yeah it works yeah it works like this so find out the
ambient to give you guys a generalization but this bracket other explanation already
okay one more point you guys know it so here's the deal let's go ahead and consider
a number that say and and I'm just going to prime factory for you guys and general we can write it
like this P1 to the alpha 1 and P1 is just a prime and Alpha One just its power so continue P2 Alpha
2 and then so so on up to PK Alpha K this has to be finite and I'm just going to say that p i r
distinct Primes just go ahead and do the prime factorization that you should wait
all right first thing first how do we find the total number of factors
of n well once we have the prime factorization just go ahead and pay attention to the exponents
what we're going to do is look at alpha one and then just add one to it and then continue look
at Alpha 2 and L1 to it and then continue right just so just put down plus times dot up to Alpha k
plus one and that's it
next let's look at the sum of factors all the factors of course some of all the factors of n
so what do we do this time we have to look at the base as well and of course the power starting from
Alpha One for the power of this right here right look at the base P1 so P1 to the zeros power and
then we add P1 to the first Power and plus up to P1 to the alpha one's power all right
next P2 for the base to the serious power first and plus P2 first plus plus P2 to the alpha 2.
and then continue and then the last one is P K to the zeros Power Plus p k to the first
plus plus p k to the alpha case power all right cool huh but you might think that hey this right
here may be kind of too hard to compute don't worry because in fact each of them is just what
of finite geometry series this right here just in case if it gets too long let's say
10 of them were 20 of them go ahead and do the following P1 to the alpha 1 plus
1 power and then minus one and then divide it by P1 minus 1. right the geometric sum formula
all right similarly this is P2 Alpha 2 plus 1 and a minus one over P2 minus one
and then so on so on and then p k Alpha K plus 1 minus one over p k minus one cool
yeah so these two are the generalization when we have to deal with the number of factors were
the sum of factors of a certain number are we done no I'm going to give you guys a little
homework question we'll talk we talk about the sum already here's the question for you guys all right
how would you find the product of factors of course all the factors of the number n ah
product of factors
of n there we go so this right here will be a question for you guys
right the hindiness just refer back to what we did right here with this
concrete example and then try to formulate a formula for this right here all right so
again this question is from the Berkeley map tournament for middle school students
go ahead and check them out I will have the link in the description for your convenience