Ambiguous Clock

The hands of my alarm clock are indistinguishable. How many times throughout the day their positioning is such that one cannot figure out which is the hour hand and which is the minute hand?

Remark: AM-PM is not important.

Imagine that you have a third hand which moves 12 times as fast as the minute hand. Then, at any time, if the hour hand moves to the location of the minute hand, the minute hand will move to the location of the imaginary hand. Therefore, our task is to find the number of times during the day when the hour hand and the imaginary hand are on top of each other, and the minute hand is not.

Since the imaginary hand moves 144 times faster than the hour hand, the two hands are on top of each other exactly 143 times between 12AM and 12PM. Out of these 143 times, 11 times all three arrows are on top of each other. Therefore, we have 2 × (143 – 11) = 264 times when we cannot figure out the exact time during the entire 24-hour cycle.

Puzzle Socks Giveaway 3

Hey, puzzlers, our friends at Soxy are offering their comfy socks in a new puzzle pattern and want to share 5 pairs with you. Solve the puzzle below, post the answer on our Facebook wall, and you can be the lucky winner of a whole set of fun socks. Click the banner below to check Soxy’s other cool items.

Last week, I got from Soxy 1 pair of brown socks, 3 pairs of brown shoes, 2 pairs of black socks, 2 pairs of black shoes, and put them all in a wooden chest. How many times should I pick a random item from the chest, so that I end up with all-matching shoes and socks to wear on Comic-Con?


Vectors -1, 0, 1

Take all 1024 vectors in a 10-dimensional space with elements ±1. Show that if you change some of the elements of some of the vectors to 0, you can still choose a few vectors, such that their sum is equal to the 0-vector.

Denote the 1024 vectors with ui and their transformations with f(ui). Create a graph with 1024 nodes, labeled with ui. Then, for every node ui, create a directed edge from ui to ui-2f(ui). This is a valid construction, since the vector ui-2f(ui) has elements -1, 0, and 1 only. In the resulting graph, there is a cycle:

v1 ⇾ v2 ⇾ … ⇾ vk ⇾ v1.

Now, if we pick the (transformed) vectors from this cycle, their sum is the 0-vector:

f(v1) + f(v2) + … + f(vk) = (v2 – v1)/2 + (v3 – v2)/2 + … + (v1 – vk)/2 = 0.