Homework problems for 5 Nov 2011:
- Let a1, a2, … be a series recursively defined as follows: a1 = 20, a2 = 11, and for n ≥ 1, an+2 is the remainder when an+1 + an is divided by 100. What is the remainder when
is divided by 8?
- (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 52011, can we get 20115 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 72011, can we get 20117 at some point?
- Let a1, a2, … be a series recursively defined: a0 = 1, a1 = 3, and for n ≥ 0, an+2 = 6an+1 – an. Prove that for each 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 =3k, then n | (2n + 1). In particular, there are infinitely many n such that n | (2n + 1). Can you find an n with at least 2 distinct prime factors such that n | (2n + 1)?
- (IMO 1971 shortlisted) Find all integer solutions to
.
- Suppose n is a positive integer such that n | (2n + 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.