Easy math puzzles

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July 23rd, 2025 at 6:03:33 PM permalink

Quote: Ace2
Your email account receives an average of one email every six minutes, following the exponential distribution. At t=0 minutes, you open your inbox and record the exact amount of time it takes to receive a new email

Starting at t=1 minute, you roll a single six-sided die every minute until you roll a 4. So the sixth roll (if necessary) will occur at t=6 minutes.

link to original post

Same scenario, but what�s the expected waiting time for:

1) the first event (email or 4) to happen?
2) both events (email and 4) to happen at least once ?

It�s all about making that GTA

SkinnyTony

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July 23rd, 2025 at 9:31:45 PM permalink

Quote: Ace2

Same scenario, but what�s the expected waiting time for:

1) the first event (email or 4) to happen?
2) both events (email and 4) to happen at least once ?
link to original post

I want to say

36/11 and 102/11

but I'm not sure

Ace2

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July 23rd, 2025 at 9:55:41 PM permalink

Quote: SkinnyTony
Quote: Ace2

Same scenario, but what�s the expected waiting time for:

1) the first event (email or 4) to happen?
2) both events (email and 4) to happen at least once ?
link to original post

I want to say

36/11 and 102/11

but I'm not sure
link to original post

Disagree. 36/11 would be the expected waiting time of rolling two dice until a 4 appears on one(or both).

102/11 would be the expected waiting time of rolling two dice until a 4 appears on one (or both), then opening the email account and adding the time needed to receive the next message. 102/11 would also be the expected waiting time of rolling two dice until a 4 appears on one (or both), then rolling a single die until a 4 appears

It�s all about making that GTA

Ace2

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July 25th, 2025 at 7:24:32 AM permalink

Quote: Ace2
Quote: Ace2
Your email account receives an average of one email every six minutes, following the exponential distribution. At t=0 minutes, you open your inbox and record the exact amount of time it takes to receive a new email

Starting at t=1 minute, you roll a single six-sided die every minute until you roll a 4. So the sixth roll (if necessary) will occur at t=6 minutes.

link to original post
Same scenario, but what�s the expected waiting time for:

1) the first event (email or 4) to happen?
2) both events (email and 4) to happen at least once ?
link to original post

Anyone working on this or should I post the answer?

It�s all about making that GTA

Ace2

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July 29th, 2025 at 9:09:29 PM permalink

Quote: Ace2
Quote: Ace2
Quote: Ace2
Your email account receives an average of one email every six minutes, following the exponential distribution. At t=0 minutes, you open your inbox and record the exact amount of time it takes to receive a new email

Starting at t=1 minute, you roll a single six-sided die every minute until you roll a 4. So the sixth roll (if necessary) will occur at t=6 minutes.

link to original post
Same scenario, but what�s the expected waiting time for:

1) the first event (email or 4) to happen?
2) both events (email and 4) to happen at least once ?
link to original post
Anyone working on this or should I post the answer?
link to original post

1) The general solution (f) is:

f = (1 - 1/e^(1/6))m + (1 + 5/6f)/e^(1/6)

This expresses the probability that an email was received within one minute times the average time needed for an email to be received (m)...assuming it was received in the first minute...plus the probability an email was not received within one minute times one minute plus 5/6 chance a 4 is not rolled and the another iteration begins

(m) is calculated by taking the derivative of 1/e^(x/6), multiplying by x and dividing by the probability of not receiving an email within one minute to get the function: x/[6e^(x/6)(1 - 1/e^(1/6)]. Then take the integral with respect to x of that function from 0 to 1 to get (m) = (6e^(1/6) - 7)/(e^(1/6) - 1) =~0.486

Plug m into the first formula and solve for f to get the answer of:

[(6e^(1/6) - 7)/(e^(1/6) - 1)*(1 - 1/e^(1/6)) + 1/e^(1/6)]/(1 - 5/6*1/e^(1/6))

=~ 3.127 minutes

2) Let p= 1/(6e^(1/6) - 5) =~ 0.479 be the probability that a 4 is rolled before an email is received. The general solution is :

f + p6 + (1-p)(6-m)

This expresses the average time needed for the first event to happen (f) plus the chance the 4 was rolled first times six (the average time for the next email to arrive is always six minutes) plus the chance an email was received first times six minus the average time it took to receive the email. Plug in f,p and m to get

[(6e^(1/6) - 7)/(e^(1/6) - 1)*(1 - 1/e^(1/6)) + 1/e^(1/6)]/[1 - 5/6*1/e^(1/6)]+ 12/(6e^(1/6) - 5)

=~ 8.87 minutes

There's a slicker way to get this answer via inclusion-exclusion. Looking at the email and die separately, it will take an average of six minutes for each to occur. Take six minutes times two minus the time spent waiting for either, so : 6*2 - f =~ 8.87 minutes

I verified both answers to three digits with numerical integrations in excel at 1/1000 spacing

Until now I cannot find a way to directly calculate the variance, though numerical integration in excel tells me it's about 8.23 for the first event to happen. The combination of discrete (die) and continuous (email) variables invalidates every calculation method I know

Incidentally, setting the average email arrival time to 1/ln(7/6) =~ 6.49 minutes would give the two events an equal chance of happening first

It�s all about making that GTA

AnotherBill

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July 29th, 2025 at 10:13:58 PM permalink

Am I the only one who feels like my head is about to explode?

SkinnyTony

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July 29th, 2025 at 10:18:25 PM permalink

Quote: AnotherBill
Am I the only one who feels like my head is about to explode?
link to original post

I'm not sure I would call this "easy". Basically if I have to break out pencil and paper I lose interest. Yes I'm lazy.

I guess I prefer the kind of puzzle that's "clever", where, when you see the trick and have the "aha" moment, the solution becomes obvious with minimal computation that you can do in your head. To me that's the difference between a puzzle and a homework question. I'm too old for homework.

Ace2

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July 29th, 2025 at 10:55:36 PM permalink

Quote: SkinnyTony
Quote: AnotherBill
Am I the only one who feels like my head is about to explode?
link to original post

I'm not sure I would call this "easy". Basically if I have to break out pencil and paper I lose interest. Yes I'm lazy.

I guess I prefer the kind of puzzle that's "clever", where, when you see the trick and have the "aha" moment, the solution becomes obvious with minimal computation that you can do in your head. To me that's the difference between a puzzle and a homework question. I'm too old for homework.
link to original post

There are several tricks and �aha� moments in that solution. Initially I thought it could not be solved until I took the derivative of 1/e^(x/6) times x then integrated from 0 to 1 to get the average waiting time for the email between 0 and 1. The inclusion-exclusion solution to #2 is one of the more clever parts of the solution

This is definitely not the first problem on this thread that pushes the boundaries of �easy�. The solutions to this one seem more complex than they really are because they combine multiple constants

It�s all about making that GTA

SkinnyTony

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July 29th, 2025 at 11:15:17 PM permalink

Quote: Ace2
Quote: SkinnyTony
Quote: AnotherBill
Am I the only one who feels like my head is about to explode?
link to original post

I'm not sure I would call this "easy". Basically if I have to break out pencil and paper I lose interest. Yes I'm lazy.

I guess I prefer the kind of puzzle that's "clever", where, when you see the trick and have the "aha" moment, the solution becomes obvious with minimal computation that you can do in your head. To me that's the difference between a puzzle and a homework question. I'm too old for homework.
link to original post
There are several tricks and �aha� moments in that solution. Initially I thought it could not be solved until I took the derivative of 1/e^(x/6) times x then integrated from 0 to 1 to get the average waiting time for the email between 0 and 1. The inclusion-exclusion solution to #2 is one of the more clever parts of the solution

This is definitely not the first problem on this thread that pushes the boundaries of �easy�. The solutions to this one seem more complex than they really are because they combine multiple constants
link to original post

Don't get me wrong, I think it's an interesting question. But definitely not easy. I'm sure that there are others in the thread that aren't easy; I didn't mean to single this question out. It was just a response to the question about the head exploding.

The rest is just my personal preference. I find computation to be tedious; I always have. This is why I always focused on abstract algebra and graph theory in school and steered clear of statistics (which I wish I had studied more because it's really useful), calculus, and anything that used numbers other than 0 and 1.

Wizard
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July 30th, 2025 at 11:34:17 AM permalink

Quote: Ace2

1) the first event (email or 4) to happen?
2) both events (email and 4) to happen at least once ?
link to original post

(6-35*EXP(-1/6)/6)/(1-5*EXP(-1/6)/6) =~ 3.605550308

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

Wizard
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July 30th, 2025 at 1:17:55 PM permalink

5*(5/6)*EXP(-1/6)/(1-(5/6)*EXP(-1/6))^2 =~40.639195

That feels too long to me, but it's what I get. I've never been much afraid of being wrong here.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

Ace2

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July 30th, 2025 at 7:09:03 PM permalink

Quote: Wizard

5*(5/6)*EXP(-1/6)/(1-(5/6)*EXP(-1/6))^2 =~40.639195

That feels too long to me, but it's what I get. I've never been much afraid of being wrong here.

link to original post

Disagree with both. I�m confident with both of the answers I posted. As mentioned, I confirmed them with numerical integrations.

Your second answer of 40.6 isn�t reasonable. If you did the events sequentially (start rolling the die after an email is received or check email account after a 4 is rolled), the expected waiting time for both to occur should be 6 + 6 =12 minutes. So if the events go concurrently, the time for both to happen must be less than that and a good estimate would be 6/2 + 6 =9, which is actually quite close to the answer of ~8.87 minutes

Last edited by: Ace2 on Jul 31, 2025

It�s all about making that GTA

Wizard
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July 31st, 2025 at 3:50:39 AM permalink

Quote: Ace2
Disagree with both.
link to original post

I was pretty confident with #1, but know I am not embarrassed to be wrong. The second I shouldn't have posted -- it was more of a progress report.

The expected time in the first minute is the integral from 0 to 1 of exp(-x/6) = 6(1-exp(-1/6))

The expected added time for the first die roll to be a 4 is exp(-1/6)*(1/6).

The probability of neither event happening is exp(-1/6)*(5/6).

Let the answer be x. Construct as:

x = 6(1-exp(-1/6)) + exp(-1/6)*(1/6) + exp(-1/6)*(5/6) * x

Then solve for x.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

Ace2

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July 31st, 2025 at 7:31:03 AM permalink

Quote: Wizard

I was pretty confident with #1

Based on what? Do you have a simulation and/or numerical integration that confirms your answer ?

It�s all about making that GTA

Wizard
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July 31st, 2025 at 9:03:17 AM permalink

Quote: Ace2
Based on what? Do you have a simulation and/or numerical integration that confirms your answer ?
link to original post

Based on the argument in my spoiler tag.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

Wizard
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August 2nd, 2025 at 11:51:11 PM permalink

There are 50 people in a line, numbered 1 to 50 sequentially. Each person can see all people numbered less than himself. Each person either always tells the truth or always lies. Everyone knows the truthfulness of everyone else.

All even-numbered people say, "Everyone in front of me is a liar."

All odd-numbered people say, "Everyone behind me is a liar."

Who is telling the truth?

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

charliepatrick

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August 3rd, 2025 at 12:40:33 AM permalink

I get that #2 and #49 have to tell the truth since they will be referring to #1 and #50, who are actually lying.

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August 3rd, 2025 at 6:15:13 AM permalink

Quote: charliepatrick
I get that #2 and #49 have to tell the truth since they will be referring to #1 and #50, who are actually lying.

link to original post

I agree!

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

Wizard
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August 3rd, 2025 at 7:32:43 AM permalink

A king has 49 gold coins. The coins weigh 1, 2, 3, ... 49 grams each. How can he divide them among seven sons so that each son gets seven coins such that the sum of the weights is equal?

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

aceside

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August 3rd, 2025 at 10:47:42 AM permalink

I find a solution by hand

[spoiler=problem 1]
1 2 4 24 47 48 49;
3 5 6 26 44 45 46;
7 11 9 22 41 42 43;
10 8 12 28 38 39 40;
13 14 17 23 35 36 37;
16 15 18 27 32 33 34;
19 20 21 25 29 30 31.

charliepatrick

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August 3rd, 2025 at 10:49:16 AM permalink

Part 1 divide some of the coins into seven piles dealing the lowest ones (1-21) from left to right then then highest ones (49-29) so that you get 1-7, 8-15, 16-21, 49-43, 42-36, 35-29 dealt out. Thus pile 1 would have 1 49 8 42 15 35; note that 1+49-50, 8+42=50, 15+35=50. Thus at the end of this phase each pile has 150, and pile 1 has the "1", pile 2 the "2" and pile 3 the "3", and 22-28 have left to be dealt out..

Part 2 The objective at this stage is to [net] add 25 to each pile using the 22, 23, 24, 25, 26, 27 and 28 coins.
So add the 28 coin to pile 3 and remove the "3" and add it with the "22". Similarly for 27/2 and 26/1. Thus the first three piles have had 25 added to them. The last four piles can each have a 25 added, either being 25, 24+1, 23+2 or 22+3.

After these two parts the piles, which had 150gm, then had [a net of] 25 grams added to each. Thus they now each weigh 175gm. Note 7*175 is the same as 1225, which is the total of 1 thru 49.

GM

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August 3rd, 2025 at 1:03:37 PM permalink

Take a 7x7 magic square (you can find one on the Wikipedia page about magic squares) and use the rows or the columns.

charliepatrick

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August 3rd, 2025 at 1:28:38 PM permalink

^ Yes that gives another way to think about a solution...

In row 1, start by writing 1 to 7 in columns 1 to 7. Then in row 2, start one to the right, and write 8 to 14 (wrapping round as necessary). Repeat this until 43-49 have been added.
The mathematics is that the entries are A, A+1, ... A+6 where A=1, B=8 etc. and each column gets +0, +1, +2... +6. so the total for each column is A+B+C+D+E+F+G+0+1+2+3+4+5+6. For instance column 1 has A, B+6, C+5...G+1.

Wizard
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August 3rd, 2025 at 4:59:53 PM permalink

Good answers and comments! Here is my solution, which was found by hand.

Coin	Son
1	1
2	2
3	3
4	4
5	5
6	7
7	6
8	6
9	4
10	5
11	1
12	2
13	3
14	7
15	7
16	6
17	3
18	5
19	2
20	4
21	1
22	1
23	2
24	3
25	4
26	5
27	7
28	6
29	7
30	6
31	5
32	4
33	3
34	2
35	1
36	1
37	2
38	3
39	4
40	5
41	7
42	6
43	7
44	6
45	5
46	4
47	3
48	2
49	1

I got the puzzle from one of my favorite YouTube channels, Mind Your Decisions. It's puzzle 6 in the prior link.

Presh goes over two methods to construct Magic Squares of odd sides.

Last edited by: Wizard on Aug 4, 2025

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

AnotherBill

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August 3rd, 2025 at 10:12:53 PM permalink

Quote: Wizard
A king has 49 gold coins. The coins weigh 1, 2, 3, ... 49 grams each. How can he divide them among seven sons so that each son gets seven coins such that the sum of the weights is equal?
link to original post

1	2	3	4	5	6	7
8	9	10	11	12	13	14
15	16	17	18	19	20	21
22	23	24	25	26	27	28
29	30	31	32	33	34	35
36	37	38	39	40	41	42
43	44	45	46	47	48	49

1 + 9 + 17 + 25 + 33 + 41 + 49 = 175
8 + 16 + 24 + 32 + 40 + 48 + 7 = 175
15 + 23 + 31 + 39 + 47 + 6 + 14 = 175
22 + 30 + 38 + 46 + 5 + 13 + 21 = 175
29 + 37 + 45 + 4 + 12 + 20 + 28 = 175
36 + 44 + 3 + 11 + 19 + 27 + 35 = 175
43 + 2 + 10 + 18 + 26 + 34 + 42 = 175

P.S.: Am I the only one whose URLs (image tags or whatever) get cut and broken when trying to post them on this forum?

Wizard
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August 4th, 2025 at 6:59:01 AM permalink

Quote: AnotherBill

P.S.: Am I the only one whose URLs (image tags or whatever) get cut and broken when trying to post them on this forum?
link to original post

Welcome to the forum. You're the first one I've heard to mention it.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

AnotherBill

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August 4th, 2025 at 7:27:25 AM permalink

Quote: Wizard
Quote: AnotherBill

P.S.: Am I the only one whose URLs (image tags or whatever) get cut and broken when trying to post them on this forum?
link to original post

Welcome to the forum. You're the first one I've heard to mention it.
link to original post

Thank you!

I sent a private message to JB about a week ago but didn't get any response.

Any URL address I try to post on the forum (in or not in tags) just gets truncated. For example, https_COLON_SLASH_SLASH_example_DOT_com_SLASH_blablabla (changed actual colons, dots, and slashes) gets truncated to _SLASH_blablabla. Checked in Opera and Firefox. OS is Ubuntu 24.04.

P.S.: Note the typical link to original post in my messages. There is no hyperlink as it should be� At least, it's not shown to me.

charliepatrick

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August 4th, 2025 at 7:43:50 AM permalink

^ I'm guessing but it could be the system's doing it.

I suspect a while ago there was a problem in that bots kept posting links to various sites or just placing spamming adverts. These used to have to be removed my moderators. So my guess is perhaps the system now automatically updates any URLs until a member reaches some threshold.

Wizard
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August 4th, 2025 at 8:08:04 AM permalink

Quote: AnotherBill

I sent a private message to JB about a week ago but didn't get any response.
link to original post

I'm curious how a new member would know about JB. Anyway, he hasn't been heard from in years. I hope he's okay.

Yes, we had to do some screening recently due to a spam attack. Maybe that has something to do with it. Also, as a new member, you can't post links, which may muddy the waters.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

AnotherBill

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August 4th, 2025 at 8:20:45 AM permalink

Quote: Wizard
I'm curious how a new member would know about JB. Anyway, he hasn't been heard from in years. I hope he's okay.link to original post

JB is the author of the "Formatting Codes" thread and is an administrator (if to believe the information provided there). That's why I decided to reach him out.

avianrandy

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August 4th, 2025 at 8:23:10 AM permalink

For the math nerds on here,in roman numerals if a line is placed above a letter it's value is increased by what?

answer given was 1,000. This surprisee as their is already for 1000 which is m. Would love to know more about this.

Dieter
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August 6th, 2025 at 2:30:57 AM permalink

Quote: AnotherBill
Quote: Wizard
Quote: AnotherBill

P.S.: Am I the only one whose URLs (image tags or whatever) get cut and broken when trying to post them on this forum?
link to original post

Welcome to the forum. You're the first one I've heard to mention it.
link to original post

Thank you!

I sent a private message to JB about a week ago but didn't get any response.

Any URL address I try to post on the forum (in or not in tags) just gets truncated. For example, https_COLON_SLASH_SLASH_example_DOT_com_SLASH_blablabla (changed actual colons, dots, and slashes) gets truncated to _SLASH_blablabla. Checked in Opera and Firefox. OS is Ubuntu 24.04.

P.S.: Note the typical link to original post in my messages. There is no hyperlink as it should be� At least, it's not shown to me.
link to original post

This is part of the anti-spam system.
New forum members are put on an initial probation period until "good standing" is established. During that probation, links and images are confounded.
This is deliberate. (It's not a bug, it's a feature.) To the best of my knowledge, no admin is able to abbreviate to probationary period.

May the cards fall in your favor.

AnotherBill

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August 6th, 2025 at 3:27:23 AM permalink

Quote: Dieter

This is part of the anti-spam system.
New forum members are put on an initial probation period until "good standing" is established. During that probation, links and images are confounded.
This is deliberate. (It's not a bug, it's a feature.) To the best of my knowledge, no admin is able to abbreviate to probationary period.
link to original post

Dieter, thank you for your explanation!

By the way, it looks like new forum members are allowed to use links in private messages. Because, when I was writing a private message to JB, I was able to insert an image.

Wizard
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August 6th, 2025 at 8:11:41 AM permalink

This video was meant to be another presentation of the boy/girl problem. However, I claim the answer presented is wrong, based on how it is phrased. I'll stop there for now and let you watch the video yourself and let you work out your own answer.

Direct: https://www.youtube.com/watch?v=cpwSGsb-rTs.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

AnotherBill

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August 6th, 2025 at 9:45:04 AM permalink

Quote: Wizard
However, I claim the answer presented is wrong, based on how it is phrased.
link to original post

Do you mean the narrator didn't say the you're 100% sure the sound was uttered by one of those frogs?

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August 6th, 2025 at 4:23:31 PM permalink

Quote: AnotherBill
Do you mean the narrator didn't say the you're 100% sure the sound was uttered by one of those frogs?
link to original post

No, that's not my issue with the wording. My issue is that not every male necessarily croaks all the time. In other words, the fact that only males croak doesn't mean every male croaks loudly enough to be heard.

To put it another way, I think you're more likely to hear croaking from two males than one male.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

AutomaticMonkey

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August 6th, 2025 at 6:32:08 PM permalink

Quote: Wizard
Quote: AnotherBill
Do you mean the narrator didn't say the you're 100% sure the sound was uttered by one of those frogs?
link to original post

No, that's not my issue with the wording. My issue is that not every male necessarily croaks all the time. In other words, the fact that only males croak doesn't mean every male croaks loudly enough to be heard.

To put it another way, I think you're more likely to hear croaking from two males than one male.
link to original post

So what you're saying is: the wording is a bit amphibious?

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August 6th, 2025 at 9:03:04 PM permalink

Quote: AutomaticMonkey
So what you're saying is: the wording is a bit amphibious?
link to original post

Ha! Couldn't have said it better myself.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

SkinnyTony

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August 6th, 2025 at 9:09:26 PM permalink

Quote: avianrandy
For the math nerds on here,in roman numerals if a line is placed above a letter it's value is increased by what?
answer given was 1,000. This surprisee as their is already for 1000 which is m. Would love to know more about this.

link to original post

The vinculum (bar) multiplies by 1000. This is different from the M, which adds 1000.

So, for example, to write the number one million, you could write 1000 Ms, or a single M with a vinculum over it.

AnotherBill