# Math Competition

1. Aug 29, 2011

### Miike012

I consider my math skills to be decent.. probably top 5% in class. Anyways, I wanted to increase my math skills by looking at old math competition tests from my school and I noticed that I am probably only comfortable with 10% of them if that. This showed me that I need to increase my algebra skills...
And obviously school text books will not help because competition tests have much harder problems and do not prepare students for problems like that... so can some one sugest something...
I really want to improve.

Last edited: Aug 29, 2011
2. Aug 30, 2011

### Zryn

Here is an old math competition test paper for Algebra (the Harvard entry examination from 1869).

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3. Aug 30, 2011

### Miike012

Those problems dont look as hard..

4. Aug 30, 2011

### Miike012

Here is an old one from 2009
http://www.mesacc.edu/~aavilez/mathclub/mathcontest/2008spring.pdf [Broken]

Last edited by a moderator: May 5, 2017
5. Aug 30, 2011

### Miike012

But are there any sources that can help me improve my algebra?

6. Aug 30, 2011

### PeterO

Very droll. Love it !

7. Aug 30, 2011

### PeterO

This test is as much for your Spatial awareness and logic skills as algebra!!

Last edited by a moderator: May 5, 2017
8. Aug 30, 2011

### Miike012

How would number 3 be solved?

9. Aug 30, 2011

### Miike012

????????????????

10. Aug 30, 2011

### PeterO

Well it would help if you were familiar with the perfect cubes

1,4,27,64,.....

many students know these up to 1000, the 10th

113 is so easy to calculate -or simple recall - as the lower powers of 11 just mimic the pascal triangle, and the rule for multiplying by 11 is so simple,

133 is easily to be shown to be bigger than 2000, so that, and any number bigger is out of the question, you can simply select the three that work.

of course 23 + 103 + 103 = 2008 but two of them are the same.

11. Aug 30, 2011

### Mentallic

Since all of the constants are even integers, we can represent them as

$$(2x)^3+(2y)^3+(2z)^3=2008$$

where x and y are some integers, thus we have a factor of 8 that we can divide by, yielding

$$x^3+y^3+z^3=251$$

And now this is much more manageable. Let's make a table of the cubes:

13=1
23=8
33=27
43=64
53=125
63=216

And evidently 73>251 so our (x,y,z) will be a combination of the first 6 integers.
Now we can either takes stabs in the dark or we can do it in a slightly more algorithmic process. If we fix one of the integers to be 6, then all we need to do is check x3+y3=251-63=35 and the sum of two distinct cubes (with each integer being up to 5) is much more manageable. Luckily, it is quite obvious that (2,3) solves this, so our (x,y,z) that solves the equation is (2,3,6) and thus a+b+c=2(2+3+6)=22.