Category Archives: brain puzzler

March 2020 Brain Puzzler Solution

Q: How many three digit perfect squares are the sum of eighteen consecutive positive integers?

A: There are 3 three digit perfect square that are the sum of eighteen consecutive positive integers: 225, 441 and 729.

If there are 18 numbers in a row, there will be 9 pairs of numbers that will add up to the same sum.  Therefore we only need to check three digit perfect squares that are divisible by 9.  Another condition is that the two middle numbers in the list have to add up to that perfect square divided by 9.

144 (12^2) is even and divided by 9 = 16 each pair would have to add up to 16.  There are no two consecutive numbers that add to 16.

225 (15^2) divided by 9 is 25 so the two middle numbers are 12 and 13.
4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 has 12 and 13 for the middle numbers and every pair going outwards 11 and 14  10 and 15 etc adds up to 25.  This list would make 9 pairs of 25 or 225.

324 (18^2) is also even like 144.  324 divided by 9 is 36 and there are no two consecutive integers that add to 36.

441 (21^2) divided by 9 is 49 so the two middle numbers are 24 and 25.
16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33

576 (24^2) is also even.  576 divided by 9 is 64 and there are no two consecutive integers that add to 64.

729 (27^2) divided by 9 is 81 so the middle two numbers are 40 and 41.
32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49

900 (30^2) is also even.  900 divided by 9 is 100 and there are no two consecutive integers that add to 100.

Notice each number divisible by 9 is a multiple of 3 squared.  The next multiple of 3 is 33 and 33^2 >1000.  Therefore, the only three numbers are:

225, 441 and 729.

 

 

 

 

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.

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)