# Canadian Senior Math Contest Question

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

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

and

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

(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.)

• For a) take n=2 and calculate. You get two equations from it. Solve this system of $a_2$ and $b_2$ – Cornman Nov 19 '17 at 1:46
• (a) is arithmetic. (b) might be induction. (c) could be Number Theory. – Gerry Myerson Nov 19 '17 at 1:47
• 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. – Ross Millikan Nov 19 '17 at 3:55
• 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. – Alexander Burstein Nov 19 '17 at 4:07