Category Archives: brain puzzler

April and May 2017 Brain Puzzler Solution

Link to the May 2017 Brain Puzzler Solution:
April’s Brain Puzzler
What is the area of the shaded region of the given 8 x 5 rectangle?

[asy]  size(6cm); defaultpen(fontsize(9pt)); draw((0,0)--(8,0)--(8,5)--(0,5)--cycle); filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8));  label("$1$",(1/2,5),dir(90)); label("$7$",(9/2,5),dir(90));  label("$1$",(8,1/2),dir(0)); label("$4$",(8,3),dir(0));  label("$1$",(15/2,0),dir(270)); label("$7$",(7/2,0),dir(270));  label("$1$",(0,9/2),dir(180)); label("$4$",(0,2),dir(180));  [/asy]

 

The area is 6.5 square units.  
Using extra lines to break the shape into triangles is a great way to solve this problem.

https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_11

March 2017 Brain Puzzler Solution

Victoria refuses to sit next to either William or Xavier. Yasmin refuses to sit next to Zack. How many ways are there for the five of them to sit in a row of 5 chairs under these conditions?

Let Victoria be V, William be W, Xavier be X, Yasmin be Y and Zack be Z.   
Scenario 1: V sits on an end seat.

Then, since W and X can’t sit next to V, that must mean either Y or Z sits next to V.
After picking either Y or Z, then either W or X must sit next to Y/Z.
Then, the last two people can be arranged in two ways.
Since there are two different end seats that V can sit in, there are a total of .

 Scenario 2: V does not sit in an end  seat.

The are 3 seats for V to choose from
In this case, then only two people that can sit next to V are Y and Z, and they
can be in either order and there are two ways to arrange W and X
 
So the total is 28

Adapted from http://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_19

January 2017 Brain Puzzler Solution

Q:  The digits one through nine are randomly arranged to make a 9 digit number.
What is the probability that the resulting number is divisible by eighteen?

A:  The number will always be divisible by 9 as the digits 1 – 9 add up to 45 (any number whose digits add up 9 or a multiple of 9 will be divisible by 9).    So every one of these 9 digit numbers will be divisible by 9.
If the number is also divisible by 2, then it will be divisible by 18.
So we need to find how often the number will be divisible by 2.
The last digit could be any number from 1-9 and 4 of those digits are even (2, 4, 6, 8) making the # divisible by 2 (and 9, see above).
So there is a 4/9 probability that the number will be divisible by 18.

BONUS: What is the probability that the resulting number is divisible by 27?
Still working on the divisibility by 27…please send comments