Category Archives: brain puzzler

August 2019 Brain Puzzler Solution

Q: On an analog clock, how many times do a clock’s hands overlap in a one week period?

A: 154
At 12:00, the hands exactly overlap but not at other times — for example, at 1:05, the hour hand has moved slightly… and at 6:30, the hour hand is halfway between the 6 and the 7.

Making the times that they will overlap only 11 times in 12 hours (60 minutes /11).

12:00, 1:05ish, 2:10ish, 3:15ish, 4:20ish, 5:25ish, 6:30ish, 7:35ish, 8:40ish, 9:45ish, 10:50ish.
In one day they will overlap 22 times so in one week, 154 times.

July 2019 Brain Teaser Solution

A wooden cube that is 20 cm on each side is composed of 1 cm x 1 cm x 1 cm  cubes.
Q1: If you paint all 6 sides of the outside of the cube blue, how many cubes will have no blue paint?
With a 3 by 3 by 3 cube, there is just one cube in the middle that would have no paint.
With a 4 by 4 by 4 cube, there are 8 cubes in the center of the cube (2 x 2 x 2) with no paint.
With a 5 by 5 by 5 cube, there are 27 cubes in the center of the cube (3 x 3 x 3) with no paint.

Answer: The pattern continues so a 20 by 20 by 20 cube, there would be 18 x 18 x 18  (5832) cubes in the center of the cube with no paint.

Q2: What if the 20 x 20 x 20 cm cube is composed of 2 cm x 2 cm x 2 cm cubes?
Answer:  There would now be 1000 cubes (10 x 10 x 10), there would be 8 x 8 x 8 (512) cubes in the center of the cue with no paint.

May 2019 Brain Puzzler Solution

Q: Let n = 2^2008 + 2008^2. What is the units (ones) digit of 2^n + n^2?

A:   This is related to the powers of two and also the last digits of squared numbers.

We first have to find n
Since the last digit of 2^(a power) has a pattern we can figure this out!

2^2008

2 to a power has a last digit pattern
2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16
2^5 = 32
2^6 = 64
2^7 = 128
2^8 = 256
2^9 = 512
2^10 = 1024
2^11 = 2048
2^21 = 2097152
2^31 =

Since it is a cycle of 4, every power divisible by 4 will end with a 6…therefore 2^2008 ends with a 6…but we actually need to know the last two digits as 2^1 and 2^11 do not share the same last digit but 2^1 and 2^21 do.  So when we think about very large exponents for powers of 2, we need to know the last two digits…
2^4 ends with 16
2^8 ends with 56
2^12 ends with 96
2^16 ends with 36
2^20 ends with 76
2^24 ends with 16
2^28 ends with 56 and so on
every multiple of 20 ends with 76
so 2^2000 would end with 76
2^2004 ends with 16
2^2008 ends with 56

2008^2 =4032064
Adding up 64 + 56 = 120 so the last two digits of n would be 20

so in evaluating 2^n + n^2
First 2^n: 2^20  or 2^120 or 2^ 220 always ends in a “6

then n^2, anything with a 0 in the ones digit squared would have a ones or units digit of zero.

Therefore, the last digit would be 6.