This question is from the 2007 Australian Mathematics cCompetition and
0% of students got this one right, although this one isn't multiple-choice
which makes it harder. So we've got this system of equations and all the variables
have to be positive integers, so whole numbers, and we gotta work out
the smallest possible value of x, so there's gonna be like multiple solutions
but we want to work out the smallest value that x can take. So you might like
to pause the video here and give that question a go.
Now personally what I like to do where you've got systems of equations like this
is just move all the variables onto one side; that's kind of like the standard
linear algebra way of handling these things. So we could say like with that
first one I might j.ust shift these over to the left
I'm just sort of more comfortable with it that way. So 3z... in
mathematics we always put a line through the z's
so they don't look like twos. So that's... I'll call that equation one. And what you also
want to do is keep all the variables in the same order, so I don't want to say x
plus z because up here I had y as the next variable so when I shift this
across I'm gonna have x minus 4y and then plus the z minus 4u. Let's
call that equation two. And then with this one we've got x minus 5y minus
5z plus y equals zero. I'm gonna call that equation three. Now kind of the
the easiest way to eliminate things would be to get rid of x's cuz like the
coefficients are all one. But since we're trying to find the minimum possible value
of x, what I'm going to do is eliminate something else instead. In the solutions
they actually did eliminate x but I found that way a bit confusing because
what they did was they they found like the minimum solutions for like y and z or
something and then they just said, okay well therefore the minimum value of x
will be when we use those values, which I didn't think was obvious. So I want to
just get rid of everything but x. All right so what I'm gonna do is... notice
that the coefficient here is 1, so that one's gonna be a good one to use to
eliminate the u's in the other equations, so like if I multiply this one by 4 and
add it to this one that's gonna get rid of the u's or if I multiply by 3 and
add it to that one it's gonna get rid of the u's there. So let's try that. Let's go
equation 1 plus 3 times equation 3 and see what happens. So one of that and 3 of
that make 4x, and then we'll have... so 1y and negative 15 will be negative 14y,
and then we'll have one of that, so negative 3 minus 15... negative 18z. The
u's will cancel and then we've got 0 on the right hand side. I'm gonna call that
equation 4. Now if any of you haven't done linear algebra before I
just might go on a tangent here about the history of this stuff. You might see
how irritating it is to keep having to write you know the x's and y's and z's
and u's and think, you know, wouldn't it be nice if we could just say, okay
that coefficient is 1 and then that coefficient is 1 and then that one's
negative 3 and that one's negative 3 and so on and wouldn't that be simpler. Well
that's exactly what the Chinese thought and I think it was The Nine Chapters on the
Mathematical Art which was one of the first examples of matrices being used
which just sort of simplifies this whole thing. Anyway... but I'm not
going to use like matrix notation here. Okay so that's equation 4; we've
gotten rid of the u. Let's do the same thing with this equation. So I'm going to
take equation 2 and add four times equation 3 because that's gonna get rid
of the u's. So we've got that plus 4, so that's 5x, negative 4 minus 20...
negative 24y, and then we'll have 1 minus 20... negative 19z, and the the
u's will cancel and we get 0. Let's call that equation 5. Now we need
to eliminate either y or z from these equations. The z's are a bit
annoying because... 19 and 18: that's gonna be irritating to multiply everything by.
But the 14 and the 24 here, we might be able to find like a lower LCM since they're
both even. So if you... the way that I think about it is that the 14 is 2 by
7, and the 24 is 2 by 12. So if I... like we don't want to repeat the two, like I
don't want to just multiply this by 14, and multiply that by 24, because then you're
sort of doubling up on the 2 which you don't need to. What we could do is just
multiply this one by 7 and this one by 12 and then that'll... the y's will cancel
out. So let's do that. So I want 12 times this one so 12
times equation 4 and then minus 7 times equation 5 and that should get rid of
the y's. Okay so we've got 12 times this, so 48, minus 7 times that, so
35, so 48 minus 35 is 13x. Those will cancel. And then, so we're not allowed
calculators for this exam so I'll just have to do 12 by 18. I don't know my 12
times tables, so I just have to do it this way. I feel like I've made a mistake
there, yes I definitely have. That was meant to be a 1. Trying to talk and
think at the same time, it's not good! Um okay 216 and then if we do 7 times the
24 then we... no! We're doing seven times 19, sorry.
7 times 19... 63. 7 times 1... 7,
and that makes 13. So we've got... that one's actually negative cuz that's
-18; that one becomes positive because we're subtracting that, so we want like
133 minus 216 and I'll just do 216 minus 133... so 83. So we'll
get... except that it's -83, because that was the one that was negative. Okay and then
what I might do is just shift that over to the right, so we've got 13x equals 83z
Now let's think about the fact that they're positive integers. So if you've
done Diophantine equations before, this might be easier for you, but the kind of
idea is that... because over here we've got, you know, 13
times an integer, that means this is a multiple of 13, right? So this has to be a
multiple of 13. Now 13 doesn't go into
83, and like they don't have any common factors; this is actually
prime. So that means that z has to be a multiple of 13, and similarly x has to be
a multiple of 83, because this is a multiple of 83, so this is a
multiple of 83, and 83 doesn't have any common factors with 13, so x
has to be, you know, 83 or two times 83 or three times 83 and so on. Now it
has to be a positive integer, can't be 0, so the smallest one, theoretically, would
be x equals 83. And that actually is the answer, but we should probably check that
that does actually work, like potentially that might result in like if we... so we've
got z = 13 and x = 83 maybe if we plug that back into here we might
get y is like 1/2 or something. So you should probably check that. I'm not going
to go through the arithmetic here but you would get, I think I got like y equals 7
and u was 17, so it does actually work. So the way that I did it was completely
different from how they did it in the solutions. They had a couple of different
methods, but I didn't really like the way that they did it because... so there was
one where they solved for like z and y and they said let's make z and y as small as
possible and then that will give the minimum solution for x and then there was
another one where they sort of worked out what x+y+z+u
was and they said, you know, minimise that and then that will minimise x. I didn't
think that either of those facts were obvious: that the smallest x would occur
when those other things are minimised, so I felt like this was the most
straightforward solution, even though it involved a bit more arithmetic than the
other ones.