# February 2020 Brain Puzzler Solution

Q: In an economy pack of 12 highlighters, 3 are defective.

If you purchase 2 highlighters (picked randomly), what is the probability that neither highlighter is defective?

A: 6/11
Since there are 3 defective highlighters out of 12, the probability of selecting the first non-defective highlighter is 9/12.
The probability of selecting the second non-defective highlighter is 8/11 as 11 highlighters remain and only 8 of them are good.
Therefore, the probability that a customer buys 2 non-defective highlighters is:
= 9/12 x 8/11
= 3/4 x 8/11
= 24/44
= 6/11.

# January 2020 Brain Puzzler Solution

Q: A: Huge shoutout and thanks to http://matchstickpuzzles.blogspot.com/

# December 2019 Brain Puzzler Solution   # November 2019 Brain Puzzler Solution In Kakuro (also known as Cross Sums), no digits can be repeated in each sum.  For example, 16 for two boxes must be 9 and 7 (or 7 and 9) and cannot be 8 and 8. # October 2019 Brain Puzzler Solution

Q:  How many factors of 155^9 are perfect squares and/or perfect cubes?

A: 37 factors of 155^9 are perfect squares and/or perfect cubes.

155 is a composite number composed of 5 x 31.
Therefore
155^9 = 5 x 5 x 5 x 5 x 5 x 5 x 5 x 5 x 5 x 31 x 31 x 31 x 31 x 31 x 31 x 31 x 31 x 31

25 Perfect Squares:
1: 1 is both a perfect square and a perfect cube.
4:  Perfect squares with just 5s = 5^2, 5^4, 5^6, 5^8
4:  Perfect squares with just 31s = 31^2, 31^4, 31^6, 31^8

Combining the 5s and the 31s Perfect Squares
4: 5^2 x 31^ 2, 5^4 x 31^2, 5^6 x 31^2, 5^8 x 31^2
4: 5^2 x 31^ 4, 5^4 x 31^4, 5^6 x 31^4, 5^8 x 31^4
4: 5^2 x 31^ 6, 5^4 x 31^6, 5^6 x 31^6, 5^8 x 31^6
4: 5^2 x 31^8, 5^4 x 31^8, 5^6 x 31^8, 5^8 x 31^8

12 Perfect Cubes:
Perfect cubes with just 5s = 5^3, 5^6 (already counted above), 5^9  (2)
Perfect cubes with just 31s = 31^3, 31^6 (already counted above), 31^9 (2)

Combining the 5s Perfect Cubes and the 31s Perfect Cubes
5^3 x 31^3, 5^6 x 31^3, 5^9 x 31^3  (3)
5^3 x 31^6, 5^9 x 31^6 (2)
5^3 x 31^9, 5^6 x 31^9, 5^9 x 31^9  (3)