0

This is abstracted from 2007 British Mathematical Olympiad Question 1.I wish to practice mathematics olympiad question for the upcoming Singapore Mathematics Olympiad Secondary 2 (Grade 8).

Find the value of $$\frac {1^4+2007^4+2008^4} {1^2+2007^2+2008^2}.$$

Note that calculators are NOT allowed to use and working MUST be provided.

For some reason,my brain immediately thought the answer is $2008\times2007+1$

I do not know if this is wrong or right.In the first place,I do not know how to begin other then knowing that $1^n$ is still 1

Jyrki Lahtonen
  • 133,153
ministic2001
  • 566
  • 1
  • 3
  • 13

2 Answers2

3

In math contests you more often than not encounter a problem you don't know how to deal with right away. My go-to approach is then to experiment. Testing how the land lies with smaller values of parameters may lead to something (if there is a cumbersome looking parameter in the problem).

Here the year $2007$ kinda sticks out. So let's see what happens, if we just look at $$ f(n)=\frac{1+n^4+(n+1)^4}{1+n^2+(n+1)^2}. $$ Pencil+paper and/or mental arithmetic gives $$ f(1)=\frac{1+1+16}{1+1+4}=3,\quad f(2)=\frac{1+16+81}{1+4+9}=7. $$ While not conclusive, this does raise the suspicion that the answer may be an integer, doesn't it? Let's expand $$ f(n)=\frac{2+4n+6n^2+4n^3+2n^4}{2+2n+2n^2}=\frac{1+2n+3n^2+2n^3+n^4}{1+n+n^2}=\frac{p(n)}{q(n)}. $$ Hmmm? Doesn't look very clean (ok, we could just crank out long division of polynomials at this point, but bear with me for a moment)?! Let's repeat the exercise and plug in some numbers. $$ \begin{aligned} p(1)&=1+2+3+2+1=9,&q(1)&=3,\\ p(2)&=1+4+12+16+16=49,&q(2)&=1+2+4=7,\\ p(3)&=\cdots=169,&q(3)&=13. \end{aligned} $$ Now our pattern recognition algorithm will kick in. It sure looks like $p(n)$ could be $q(n)^2$. Let's try and prove that's the case $$ q(n)^2=(1+n+n^2)^2=(1+n+n^2)+n(1+n+n^2)+n^2(1+n+n^2)=\cdots=p(n). $$ We have a winner!

I'm not implying that this is the way to solve this specific problem. It is just a route that occured to me. Don't know if I would have come up with this in grade 8? Can't go back in time. Experimenting did show me the way with a couple of IMO problems back in the day. The path is not straight and narrow - you find your own!

Jyrki Lahtonen
  • 133,153
  • I was in dilemma whether I should write $2007$ as $(2008 - 1)$ or $2008$ as $(2007 + 1)$. But I like the identity you got here. +1 for it. – Bhaskar May 09 '15 at 09:05
  • This is quite late but you can divide that polynomial and get a clean cut polynomial. – Ian L Sep 26 '16 at 15:56
  • Yes, you can @IanLimarta - I even said so. I don't know whether those contestants are familiar with polynomial (long) division? The contestants probably are, but I was trying to lead up to it even in case they aren't. – Jyrki Lahtonen Sep 26 '16 at 16:43
1

Write $2007$ as $(2008 - 1)$ in both numerator and denominator and expand them using binomial theorem to get:

$\frac{1^4 + 1^2 * 2008^2 + 2008^4}{1^2 + 1*2008 + 2008^2} = 1^2 - 1*2008 + 2008^2 = 1 + 2007*2008$

[using identity $(a^4 + a^2b^2 + b^4) = (a^2 + ab + b^2)(a^2 - ab + b^2)$]

Bhaskar
  • 458