Shady and Shadowy

Math: Finding the Shortest Grid Paths Possibilities

This is a brain teaser. The rule is we are only allowed to use the the grid line (the border of each box) to reach the destination . Finding the Shortest Path Let's right away find the shortest path to reach the destination. We need to pick only two directions from all four directions: up-down-right-left. And the two must be perpendicular to each other. If we start from the top left, to go diagonally to the bottom right, we will have two options to move, those are: down — right . Other than those, we won't be going on the shortest path. If we start from the bottom left, and we have to reach the top right, then we'll have two options, those are up — right . I put different colors, green and red, so you can see them. In that example, we have 6 steps as our shortest path. How to Actually Get the Shortest Route (Steps)? By counting along the side borders of the grid. Or Adding the total rows and columns. Let's generalize the method....

Math: Finding GCF and LCM Using Continuous Division (Translated to JavaScript)

Hi! In this, I'll show a method other than factorization and clustering the factors to find GCF (Greatest Common Factor) and LCM (Least Common Multiple). I assume you're familiar with the first method. That is using tree factors and then grouping the dilly. I was taught to do this method in elementary school. The method has already been translated to an application on Port Raptor: GCFLCM.C . There's an explanation about the method flow on that post too. Let's start to implement the continuous division method. For example, we want to find GCF and LCM from: 8, 16, 18 Method: Create a table and write down those numbers as first row. Give some space for vertical column on the leftmost. The leftmost column will be used to list the divisor. The cells between the first row and first column are the division results (quotients). The division goes diagonally. Start with the division with 1, well... Then, divide starting from the lowest prime number → 2. If an e...

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