**Homework problems for 5 Nov 2011:**

- Let
*a*_{1},*a*_{2}, … be a series recursively defined as follows:*a*_{1}= 20,*a*_{2}= 11, and for*n*≥ 1,*a*_{n+2}is the remainder when*a*_{n+1}+*a*is divided by 100. What is the remainder when is divided by 8?_{n} - (IMO 1962 Q1) Find the smallest positive integer with the following properties:
- in decimal representation, its last digit is 6;
- if we move this digit to the front of the number, we get 4 times the original number.
- Find two 3-digit numbers (
*abc*) and (*def*) such that upon placing them side-by-side, we get a 6-digit number (*abcdef*) satisfying : (*abcdef*) = ( (*abc*) + (*def*) )^{2}. - (a) A positive integer is written on the board. At each turn, we erase its unit digit and add 4 times that digit to the remaining number. Starting from 5
^{2011}, can we get 2011^{5}at some point? (b) Suppose instead, at each turn, we erase its unit digit and add 5 times that digit to the remaining number. Starting from 7^{2011}, can we get 2011^{7}at some point? - Let
*a*_{1},*a*_{2}, … be a series recursively defined:*a*_{0}= 1,*a*_{1}= 3, and for*n*≥ 0,*a*_{n+2}= 6*a*_{n+1}–*a*. Prove that for each_{n}*n*, either or is a perfect square.

**Homework problems for 12 Nov 2011:**

- Prove that if
*a*,*n*are integers and*n*> 0, then whenever , we have . - Classify the solutions of , where
*x*,*y*and*z*are positive integers. - Prove that if
*n*=3^{k}, then*n*| (2^{n}+ 1). In particular, there are infinitely many*n*such that*n*| (2^{n}+ 1). Can you find an*n*with at least 2 distinct prime factors such that*n*| (2^{n}+ 1)? - (IMO 1971 shortlisted) Find all integer solutions to .
- Suppose
*n*is a positive integer such that*n*| (2^{n}+ 1). Prove that*n*= 1 or*n*is a multiple of 3.

[ Coming attractions : quadratic residues… stay tuned! 🙂 ]

Homework problems for 12 Nov 2011: Q5

something wrong with this question, if n=6, (multiple of 3), but n cannot divide 2^{n}+1.

n should be a power of 3, i.e, n=3^{k}, k\in \mathbb{N}_{0}

The implication is only one way: the converse doesn’t hold.