I was trying to solve the problem below but was completely unable to. My question is that which general category do questions like these belong to?. Once, I know that I would go learn the specific math concept and attempt at solving the question.

For each positive integer $n$, define $a_n$ and $b_n$ to be the positive integers such that

$$\left(\sqrt 3 + \sqrt 2\right)^{2n} = a_n + b_n\sqrt 6$$


$$\left(\sqrt 3 - \sqrt 2\right)^{2n} = a_n - b_n\sqrt 6$$

(a) Determine the values of $a_2$ and $b_2$.

(b) Prove that $2a_n - 1 < \left(\sqrt 3 + \sqrt 2\right)^{2n}<2a_n$ for all positive integers $n$.

(c) Let $d_n$ be the ones (units) digit of the number $\left(\sqrt 3 + \sqrt 2\right)^{2n}$ when it is written in decimal form. Determine, with justification, the value of $d_1+d_2+d_3+\cdots+d_{1865}+d_{1866}+d_{1867}$ (the given sum has 1867 terms.)

  • $\begingroup$ For a) take n=2 and calculate. You get two equations from it. Solve this system of $a_2$ and $b_2$ $\endgroup$ – Cornman Nov 19 '17 at 1:46
  • 2
    $\begingroup$ (a) is arithmetic. (b) might be induction. (c) could be Number Theory. $\endgroup$ – Gerry Myerson Nov 19 '17 at 1:47
  • 1
    $\begingroup$ It is important that $\sqrt 3 - \sqrt 2 \lt 1$ so when you raise it to a power it gets small. Then note that $(\sqrt 3 + \sqrt 2)^{2n}+(\sqrt 3 - \sqrt 2)^{2n}$ is an integer because the terms with square roots in them cancel. This is a recurring theme in olympiad problems. $\endgroup$ – Ross Millikan Nov 19 '17 at 3:55
  • $\begingroup$ Look at the units digits of both $a_n$ and $b_n$ for small $n$ (say, through $n=5$) and see if you notice anything useful. $\endgroup$ – Alexander Burstein Nov 19 '17 at 4:07

I would call all of these algebra, but this is definitely competition math algebra. These are solvable by expanding through binomial expansion (and I believe this is the intended solution).

If you need more hints, ask below.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.