The Puzzler – Week 10

Welcome to The Puzzler. Every week, there are 2 new puzzles related to my passion: math, logic, and thinking. The first puzzle will be The Puzzler Quik, meant for those who crave something fun-sized. The second puzzle will be The Puzzler Think, meant for those who love to ponder. The answers will be posted in next week’s column. Don’t forget to submit your answer for a potential shout-out in the next edition of the Puzzler.

 

Don’t forget to check out The new article in our Mind Bogglers Section. It features a traveling problem that just isn’t possible

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Are you interested in math and puzzles? Do you like winning money? Well, why not participate in ThePuzzlr Contest, a 40-minute, 30 questions test for all ages that will be available to take via our website! We even have a sample test to help you get a feel of the problems! If you have already signed up, be sure to share this with as many friends as you can.

Puzzler Quik

Grandma left half her money to her granddaughter and half that amount to her grandson. She left a sixth to her brother, and the remainder, $1,000, to the dogs’ home. How much did she leave altogether?

Puzzled?

Let x= the amount of money that she left altogether. Can you make a equation

You don’t need a second hint!

Puzzler Think

In how many ways can we color the $8$ vertices of an octagon each red, green, or blue, so that no two adjacent vertices are the same color?

Puzzled?

Is  there a way to fix the last color?

Last Week's Puzzler Quik

Joshua from Toronto, Canada was the winner of this shout-out. His explanation was superb!

Each player plays twice against members of other teams. There are 2 player in a team and each player plays against the other 4  players twice. Also, each player plays exactly one game with his teammate. This gives us a total of $$8+1=9.$$ games. Since there are 6 players, we have a total of 6*9=54 games. Since each game is played by 2 players, we get $$\frac{54}{2}=27.$$ Thus, 27 games were played.

Last Week's Puzzler Think

This week’s shoutout goes to Nehal from Illinois! Kudos to him for solving this problem!

Here is a way to do this problem:

Let’s expand Masha’s rewritten quadratic. $$(x+m)^2-63=x^2+2mx+(m^2-63).$$ We know that the units term in the original quadratic is 1, so m^2-63=1 and m^2=64. This means that m is positive or negative 8. Then, we can determine that 2m=b, since they are both the linear coefficients in each form of the quadratic. We can easily determine that m and b have the same sign, thus m=+8, and b=16.

Puzzled? Questions? Suggestions? Holler at aditya@thepuzzlr.com

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