Easy math puzzles
Poll
 | 25 votes (49.01%) | ||
| | 16 votes (31.37%) | ||
| | 7 votes (13.72%) | ||
| | 4 votes (7.84%) | ||
| | 12 votes (23.52%) | ||
| | 3 votes (5.88%) | ||
| | 6 votes (11.76%) | ||
| | 5 votes (9.8%) | ||
| | 12 votes (23.52%) | ||
| | 10 votes (19.6%) |
51 members have voted
February 26th, 2026 at 4:25:45 PM
permalink
Let's say that you start with 15 ml of milk and 250 ml of coffee. You move one teaspoon (5 ml) of milk into the coffee mug.
The ratio of coffee to total volume becomes 250/255 which is about 98%. Then, you take one teaspoon (5ml) of the 98% coffee + 2% milk mixture and dump it into the milk container. This teaspoon is 98% coffee and 2% milk, so .98*5ml coffee = 4.9 ml and 0.1 ml milk.
The milk container had 10 ml of milk after a teaspoon of it was removed. After adding back 5 ml, its volume is again 15 ml, and most of that, but not all, that is returned (98%) is coffee.
The amount of coffee in the milk cup is 4.901961 ml. The amount of milk in the coffee cup is 5 ml.
The ratio of coffee to total volume becomes 250/255 which is about 98%. Then, you take one teaspoon (5ml) of the 98% coffee + 2% milk mixture and dump it into the milk container. This teaspoon is 98% coffee and 2% milk, so .98*5ml coffee = 4.9 ml and 0.1 ml milk.
The milk container had 10 ml of milk after a teaspoon of it was removed. After adding back 5 ml, its volume is again 15 ml, and most of that, but not all, that is returned (98%) is coffee.
The amount of coffee in the milk cup is 4.901961 ml. The amount of milk in the coffee cup is 5 ml.
February 26th, 2026 at 7:22:44 PM
permalink
Quote: KevinAALet's say that you start with 15 ml of milk and 250 ml of coffee. You move one teaspoon (5 ml) of milk into the coffee mug.
The ratio of coffee to total volume becomes 250/255 which is about 98%. Then, you take one teaspoon (5ml) of the 98% coffee + 2% milk mixture and dump it into the milk container. This teaspoon is 98% coffee and 2% milk, so .98*5ml coffee = 4.9 ml and 0.1 ml milk.
The milk container had 10 ml of milk after a teaspoon of it was removed. After adding back 5 ml, its volume is again 15 ml, and most of that, but not all, that is returned (98%) is coffee.
The amount of coffee in the milk cup is 4.901961 ml. The amount of milk in the coffee cup is 5 ml.
link to original post
I'm not sure where your mistake is, but here are the amounts and ratios after each movement.
After first movement
| Cup | Coffee | Milk | Total | Ratio Coffee | Ratio Milk |
|---|---|---|---|---|---|
| Coffee | 250 | 5 | 255 | 0.980392 | 0.019608 |
| Milk | 0 | 10 | 10 | 0.000000 | 1.000000 |
| Cup | Coffee | Milk | Total | Ratio Coffee | Ratio Milk |
|---|---|---|---|---|---|
| Coffee | 245.0980392 | 4.901960784 | 250 | 0.980392 | 0.019608 |
| Milk | 4.901960784 | 10.09803922 | 15 | 0.326797 | 0.673203 |
Note the milk in the coffee cup = coffee in the milk cup = 4.901960784.
We at least agree there is 4.901960784 ml of coffee in the milk cup. That had to displace the same amount of milk that used to be there. Where else could it have gone but the coffee cup?
"My life is spent in one long effort to escape from the commonplace of existence. These little problems help me to do so." -- Sherlock Holmes
February 26th, 2026 at 8:54:27 PM
permalink
OK, I get it now. That was a good one!
February 27th, 2026 at 7:17:34 AM
permalink
You flip a fair coin 50 times. What is the probability of never seeing three or more heads in a row? A numeric answer to six significant digits will suffice.
"My life is spent in one long effort to escape from the commonplace of existence. These little problems help me to do so." -- Sherlock Holmes
February 27th, 2026 at 10:21:16 AM
permalink
Exact answer should be the 52nd tribonacci number divided by 2^50. There is a closed form formula for tribonaccis but it’s much longer than the Fibonacci formula.
Very easy to Markov the answer though I consider that brute forcing it
Very easy to Markov the answer though I consider that brute forcing it
It’s all about making that GTA
February 27th, 2026 at 10:32:28 AM
permalink
This was about the 500th puzzle on this thread that wasn’t “easy”Quote: WizardI've been working on this on my own. I found this video, by one of my favorite YouTube channels, to be very useful in finding the nth term in the Fibonacci sequence. From there, I can see how to adapt the logic for the dice question.
Direct: https://www.youtube.com/watch?v=ITSbuT9ojOw
By the way, this was hardly an Easy math puzzle.
link to original post
It’s all about making that GTA
February 27th, 2026 at 10:38:59 AM
permalink
Quote: Ace2Exact answer should be the 52nd tribonacci number divided by 2^50. There is a closed form formula for tribonaccis but it’s much longer than the Fibonacci formula.
link to original post
It does not help that determining a closed form for a recursive sequence that depends on the three previous terms requires finding the roots to a cubic equation.
February 27th, 2026 at 11:28:41 AM
permalink
I'm still looking for a simple Markov Chain answer, in part to verify my own answer.
"My life is spent in one long effort to escape from the commonplace of existence. These little problems help me to do so." -- Sherlock Holmes
February 27th, 2026 at 1:13:45 PM
permalink
Here are some riddles for you.
- What has a face and hands but no arms or legs?
- If I have a bee in my hand, what is in my eye?
- What's harder for you to catch the faster you run?
- I appear twice in the morning. I appear twice in the evening. But I only appear once at night. What am I?
- What has 13 hearts but no other organs?
- What has three feet but can't walk?
- What has a tail but no body?
- What is so delicate that if you say its name, it breaks?
- What is always on its way but never arrives?
- Five friends are together in a room. Charlie is knitting. Don is cooking. Ace is playing chess. Gordon is reading a book. What is the fifth friend doing?
"My life is spent in one long effort to escape from the commonplace of existence. These little problems help me to do so." -- Sherlock Holmes
February 27th, 2026 at 1:48:21 PM
permalink
What has a face and hands but no arms or legs?
An analog clock
If I have a bee in my hand, what is in my eye?
What's harder for you to catch the faster you run?
Your breath
I appear twice in the morning. I appear twice in the evening. But I only appear once at night. What am I?
The letter N
What has 13 hearts but no other organs?
A deck of cards
What has three feet but can't walk?
A yardstick
What has a tail but no body?
A coin (usually)
What is so delicate that if you say its name, it breaks?
Silence
What is always on its way but never arrives?
Tomorrow
Five friends are together in a room. Charlie is knitting. Don is cooking. Ace is playing chess. Gordon is reading a book. What is the fifth friend doing?
Watching one of those videos on Facebook where you click on the link and it brings up a page loaded with ads but it never does reveal the answer to what was on the original video
February 27th, 2026 at 2:55:05 PM
permalink
Very simple Markov chain with only three states (last 2, 1, or 0 flips were heads). The sum of the three states after fifty iterations is the answer…gives you the probability that there were never more than two consecutive headsQuote: WizardI'm still looking for a simple Markov Chain answer, in part to verify my own answer.
link to original post
It’s all about making that GTA
February 27th, 2026 at 3:35:02 PM
permalink
I haven't completed the spreadsheet but here is a possible approach. It considers fifty empty slots which are to be filled with "H" or "T", in such a way that there aren't any "HHH"..
Consider three kinds of lines
(a) lines which end "HH"
(b) lines which end "H"
(c) lines which do not have a "H" in last position
Then, initially, fill the remainder of the line with tails.
Now consider two kinds of additions
(i) "HT" (one head)
(ii) "HHT" (two heads)
Note for the each series of "Heads" to finish, there must be room for a "Tail" after it/them. As we're only interested in one or two Heads, then these are the only two types of additions allowed. Since there is a possiblity, at the end of a line, for none, one or two Heads not followed by a Tail, then looks at these types of lines - which now have 48,49,50 "empty" slots.
For each type of line how many of (i) or (ii) can one add
"HH" will have 48 slots, "H" will have 49, "none" will have 50
Another idea is to use similar logic for (say) a row of ten, identifying how many start or end with any Heads (i.e. whether they could be put next to each other). (I might look at this a bit more later!)
Consider three kinds of lines
(a) lines which end "HH"
(b) lines which end "H"
(c) lines which do not have a "H" in last position
Then, initially, fill the remainder of the line with tails.
Now consider two kinds of additions
(i) "HT" (one head)
(ii) "HHT" (two heads)
Note for the each series of "Heads" to finish, there must be room for a "Tail" after it/them. As we're only interested in one or two Heads, then these are the only two types of additions allowed. Since there is a possiblity, at the end of a line, for none, one or two Heads not followed by a Tail, then looks at these types of lines - which now have 48,49,50 "empty" slots.
For each type of line how many of (i) or (ii) can one add
"HH" will have 48 slots, "H" will have 49, "none" will have 50
| "HT" | "HHT" | check | 50 slots | 49 slots | 48 slots |
| 0 | 0 | 0 | 1 | 1 | 1 |
| 0 | 1 | 3 | 48 | 47 | 46 |
Another idea is to use similar logic for (say) a row of ten, identifying how many start or end with any Heads (i.e. whether they could be put next to each other). (I might look at this a bit more later!)
