i know i graduated from uc berkeley as an undergrad
but i just really really like the stanford math tournament (SMT) for high school students
because they really have some fun math questions go ahead and check them out i will have the link
in the description for your convenience today we are going to do the algebra tiebreaker in 2021
we have three questions and we have to finish them in 15 minutes so let's get work first we are going
to find all the integer solutions to this equation and we see that we have the n here and n here
but we see these two logs they have different bases which is not a good idea so let's change it
change the water well we see that we have an eight and one over four so maybe it's a good idea to
work with log base 2. so let's do that so rewrite this as log base 2 of n over log base 2 of a
and then do the reciprocal so we will actually have log base 2 of 8 over log base 2 of n and then
this right here same thing but we will have log base 2 of n on the top over here log base 2 of 1
over 4 and this is equal to negative 5 over 2. all right this right here is just equal to 3. one way
to do it is look at the 8 as 2 to the 3rd power and then cancel the log base 2 and also the 2
right so we just have 3 and then over this is log base 2 of n plus here we have log base 2 of n over
write this down as 2 to the negative 2 and then cancel the bases so we get negative 2 here and
then this is equal to negative 5 over 2. okay now we see that we have log base 2 of n here and also
log base 2 and here so it's a good idea to just multiply everybody by log base 2 of n so let's
go ahead and do that and perhaps also multiply the two so now we don't have the denominator
so multiply two log base two of n so when we do this multiply by that they cancel we will get six
and then this that can so we have the minus but this and that together will give us parentheses
with log base two of uninside and then square and that's equal to the twos cancel and we have
the minus five log base 2 of n yeah so far so good and now this right here actually becomes
a quadratic equation by in terms of log base 2 of n so we're going to change the water a little bit
put this and that to the right hand side but i'm going to write them down first
so we will get parentheses with log base 2 of any ciphers and then square that
minus 5 log base 2 of n and then minus 6 and that's equal to 0. okay we can factor this
which is very nice factor in terms of log base 2 of n so we just need this right here and then
another parenthesis with log base 2 of n cool we need to produce negative 6 so we need negative 6
and positive 1 so let's say we have minus 6 here and then plus 1 here so this means log base 2 of n
6 is equal to 0 means that this right here should be equal to plus 36 or the other possibility
is log base 2 of n equals negative 1. okay to solve this what we can do is just 2 to the power
this this they cancel and then 2 to this 2 to that they cancel so what do we get
n equals 2 to the 6th power is 64. yes and then this right here n is equal to
two to a negative one which is one half but the question is asking integer solutions only so we
don't want this the only answer that we want is positive 64. number two it's a pretty weird one
it took me a while to understand what the question is asking so let's have a look it says let f be
a function that goes from end to end and we want the function to be a bit reaction so that for oa
either we get a is equal to f of a squared or f of a is equal to a squared the question is asking for
how many numbers less than 40 is f of a not equal to a squared so what's the question asking right
let me break it down for you first off this is a steady bad notation because we can just take the
square roots on both sides and just a positive square root because n is just a set of all the
positive whole numbers so i'm going to rewrite this as f of a equals square root of a all right
and you might be wondering do we include 0 for n well let me tell you it doesn't matter
for this question and in general it depends on your convention and depends on the agreement
that's all so i'm just going to say no and let me just write down some natural numbers for you guys
and the next one is 10 but i'm just going to skip that because we are talking about square root of a
and we are just talking about whole numbers we just want to make sure to look at some perfect
squares so i will jump to the next perfect square which is 16 and then again i'll jump to the next
perfect square which is 25 and then 36 and then i'm just going to you know write it down 37 well
i'm just going to end up with 39 because we don't need to know 41 more right so that's what we'll
be looking at and this is like the set of the domain and now i'm going to write this down again
so this is how i was written out the question so check this out in order for a function to be
a projection we just have to make sure that we have to assign a number from here to a number
here so they will just have to pair up and you can have any duplication for example you cannot assign
four and five to both equal to five that wouldn't be a direction
it has to be one to one and hung two all right let's get to work the first number is one what
do we do do we do the square root of one what do we do one square same thing right one square
and square root of one same thing they give you one so it doesn't matter it must go to one good
next we are looking at 2 what do we do can we do square root of 2 no because square root of
2 is not possibility here so we must do 2 squared so we will have to assign the 2 to 4
all right and then similarly for 3 we cannot assign that to square root of 3 so we must
square that so this right here will be assigned to here and then check this out for the number 4
we don't want to square that to 16. do we really want to do that no
why because we haven't used the 2 yet here we must assign the 4 to 2 because there's no other numbers
that can possibly be assigned back to 2 here so we must do the square root of 4 to get the 2 here
all right next for the 5 we'll just square that which is going to be
here and then for the 6 we'll square that to be here and then for 7 it's going to be 49 i'm
just going to draw the arrow like this because i didn't write down 49. similarly for 8 square that
now for 9 what do we do we will have to do the square root of nine yeah so same reason we will
have to assign the nine to three because again you want to make sure that these guys are taken
cool and then 10 11 up to 15 we'll just have to square them which is going to be
here and then let's consider 16 what do we do do we assign 16 to 16 square
yes because you see that the 4 is taken already
even though this 16 is a perfect square like 9 9 4 but if we take the square root of 16
this right here is taken already so we cannot assign 16 to 4. cannot do that we will have to
assign the 16 to 16 squared all right and then if you look at 25 what do we do we will take the
square root of 25 and then we will have to assign that to here and that's how we can cover the five
all right and then if you look at 36 what do we do do we assign far away what do we ascend to the
six this is not taken yet so we will just have to assign oops wrong color not better take the square
root of that so it's here now cool and that's pretty much it so we're talking about 4 5 25 and
36 these numbers for a we will do the square root and you see we didn't square that so the answer
for this question is four and these four numbers are 4 9 25 and 36. okay last question and in fact
this is my favorite one as well let's see let x be a real number satisfying this quadratic equation
and if this is true then this must also be true for some number k and our goal is to find out
what k is does that mean we have to solve this quadratic equation by using a quadratic formula
no we don't have to in fact this is a pretty classic algebra map composition question
whenever you have a quadratic equation like this and you want to enter attack a bigger exponent
we can do the following check this out let me write this down again x squared minus 3x plus
1 is equal to 0. and if you have never done this question before maybe you will attempt to move the
1 to the other side and then start to square both sides so you can get a bigger and bigger power
right the problem of doing that is we will end up with more and more terms on the left hand side
but don't worry this is how we can do it let's divide everybody by x
including this this part and we will get x minus 3 plus 1 over x and that's equal to 0 and then
move the minus 3 to the right hand side so we are going to get x plus 1 over x is equal to 3.
why did we do that well check this out now let's square both sides when we do this
we will just do the first thing square which is x squared and we add
2 times this and that what's x times 1 over x it's just 1. so 2 times 1 is just 2.
it's just a number and then we'll square this so we add 1 over x squared and that's equal to 9.
well you see that even though we have three terms but this is just a number so we can move the two
to the other side so in fact we'll just get x squared plus one over x squared equals seven
and guess what yes we can repeat the same process let's go ahead and square this again how many more
times will just continue let's see we will have to square that which is x to the fourth power
and we add two times this and that which is just a two and then we do this square which is
one over x to the fourth power and we get fourteen not just kidding forty-nine move the two to the
other side so we get x to the fourth power plus one over x to the fourth power equals 47 and
then we square both sides okay so we get x to the eighth plus 2 times this and that which is just 2.
so nicer and 47 squared i looked up the answer earlier already it's 2 2 0
9. so i didn't do this in my head okay anyway move the two to the other side we are getting
x to the a plus one over x to the eighth that's equal to two two zero nine seven sorry
should we square again because the power here is 16
no we don't have 2 because this is 1 over x to the 8th here we will just multiply
everybody by x to the 8th this times this gives us the x to the 16. this times that is 2 2 0 7
x to the 8th on the right hand side let's move it to the left hand side so we will have a minus 2 2
0 7 and then we have x to the 8th and then this times that is just plus one and that's equal to
zero better all right as you can see this is exactly what the question told us so what's the k
of course k is equal to positive 2 2 0 7 because they put a negative right here so k is plus d
so that's it this is a very very classic algebra math competition equation for such setup stanford
math tournament is definitely one of my favorite math competition type of graphic questions of
course i have algebra and also calculus and some probabilities and you guys can also try the other
categories and let me know how it goes if you guys want to see another standard math
questions you guys can check out this video where i rotate the parabola i'll see you guys over here