Showing posts with label difficulty-1. Show all posts
Showing posts with label difficulty-1. Show all posts

Saturday, December 6, 2014

Problem Post 2-4: On the Escalation of P*cking Problems

Problem Posts
Image Credit: "Cardboard Box City" by James Nash. Licensed under CC BY-NC-SA 2.0.
No, I don't think they'll make you solve this problem in fifteen seconds mentally.

You may wonder why I used the P-word for something as "simple" as packing boxes. I mean, the question is straightforward... right?

THE PROBLEM
You have a big rectangular pallet upon which you must place identical rectangular boxes. You can't stack, collapse, or overlap boxes. The sides of the boxes have to parallel to one of the sides of the pallet. What's the maximum number of boxes you can place?

It's a familiar but unremarkable question, especially for many mathletes who've had to do it under time pressure. (Fifteen seconds mental? Easy!) And the real-life applications are obvious - putting things into shipping containers, for example.

Friday, September 26, 2014

Problem Post 2-1: Factor Sums, and the Distributive Law

Problem Posts
Who knew power series multiplication could help you at the grocer? (see Question 2)

Just a quick post for younger readers. Often one will find oneself using `brute force' approaches when obvious tricks and shortcuts exist. Most of the time, this is justified -- many tricks are usually too arcane to remember, or too impracticable to execute realistically. This is neither. It's fast, simple, and it could save you in a Do-Or-Die (I've used it before; our team won!)

What is a factor?
First things first, right? Not everybody defines the word “factor” the same way. I will be using the convention used in most local contests: A factor of an integer $n$ is a positive integer $a$ for which there exists an integer $b$  such that $n=ab$. So by our convention $4$ and $-4$ both have exactly three factors: 1, 2, and 4. We do not consider $-2$ as a factor, even if it divides both $4$ and $-4$. On the other hand, a factor of a number is a proper factor iff it is not the number itself. Often 1 is also not considered a proper factor. With this ambiguity, however, the term is not used too often in contests.

Tuesday, December 3, 2013

Problems 1-4 Mass Points

Problem Posts
Problem 1-4 (Geometry) [Difficulty 1]

QUESTION
(Classic) $D$ bisects side $AB$ of triangle $ABC$. $E$ is on $AC$ so that $AE:EC=1:2$. $CD$ and $BE$ meet at $G$. Find $DG:GC$.

Overview
Mass Points.

Saturday, November 16, 2013

Problem 1-3 Coin Rotation

Problem Posts
Problem 1-3 (Geometry) [Difficulty 1] [PDF]

QUESTION
(Original) Three identical cylindrical barrels with radius $\sqrt{3}$ are placed tangent to each other (represented by circles $c_{1}$ , $c_{2}$ , $c_{3}$ .) A metal sheet $AB$ is placed just touching $c_{1}$ and $c_{3}$. A coin $c$ with radius $2-\sqrt{3}$  is placed on the floor tangent to $c_{2}$ and $c_{3}$  (see the diagram), and rolled without slipping about the barrels (so that the coin is rotating clockwise), going through plank $AB$, until it returns to its starting point. How many radians has the coin rotated? (For example, half a turn is $\pi$ .)

Overview
Separate the coin's rotation into two 'components'.

Sunday, November 10, 2013

Problem 1-1 Kookie's Cookies



Problem Posts

Problem 1-1 (Number Theory) [Difficulty 1] [PDF]

QUESTION
(Sipnayan 2012 High School Final Round: Weightlifting 4) Kookie has a kooky way of eating cookies. He lays them out on a circle. After Kookie eats a cookie, he skips the next (clockwise) remaining cookie in the circle and eats the next (clockwise) remaining cookie after that. Kookie places a batch of $2012$ cookies numbered $1, 2, \ldots 2012$ in that clockwise order, and begins to eat them, cookie $2$ first. Let $k$ be the number on the last of the $2012$ cookies that he eats. Kookie, unsatisfied, arranges another batch of $k$ cookies numbered $2, 4, 6, \ldots 2k$ in that clockwise order, and begins again with the cookie numbered $2$. What is the number on the last cookie Kookie eats from this batch?


Overview
This problem comprises two iterations of the Josephus problem.