Easy math puzzles
Poll
 | 21 votes (45.65%) | ||
| | 14 votes (30.43%) | ||
| | 6 votes (13.04%) | ||
| | 3 votes (6.52%) | ||
| | 12 votes (26.08%) | ||
| | 3 votes (6.52%) | ||
| | 6 votes (13.04%) | ||
| | 5 votes (10.86%) | ||
| | 12 votes (26.08%) | ||
| | 10 votes (21.73%) |
46 members have voted
May 21st, 2025 at 7:17:21 AM
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Ace2,
So would 22, 44, 66 qualify, since they have 11 in common? Your calculation seems to preclude this set, since they also share a common factor of 2.
Dog Hand
So would 22, 44, 66 qualify, since they have 11 in common? Your calculation seems to preclude this set, since they also share a common factor of 2.
Dog Hand
May 21st, 2025 at 11:10:26 AM
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Good catch. One must be very precise when using sets/logical arguments/inclusion-exclusion. I modified my answer below (in caps).Quote: DogHandAce2,
So would 22, 44, 66 qualify, since they have 11 in common? Your calculation seems to preclude this set, since they also share a common factor of 2.
Dog Hand
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Therefore, the chance that three random integers share a MINIMUM factor >10 is 0.1681 - 0.1666 = 0.15%.
Put another way, over 99% (1666/1681) of SETS WITH shared factors will INCLUDE 2,3,5 or 7.
It’s all about making that GTA
May 22nd, 2025 at 5:46:58 PM
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My conjectures on the average gap between prime numbers are:
ln(x) is the average gap at x.
ln(x) - 1 is the average gap from 1 to x
ln(x) * (pi)^.5/2 is the standard deviation of the average gap at x
ln(x) is the average gap at x.
ln(x) - 1 is the average gap from 1 to x
ln(x) * (pi)^.5/2 is the standard deviation of the average gap at x
It’s all about making that GTA
May 28th, 2025 at 11:02:23 AM
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It takes five days to paddle a canoe down a river with the current.
It takes seven days to paddle a canoe up the same river against the current.
1. How long would it take with no current?
2. How long would it take to get down the river with the usual current without a paddle?
It takes seven days to paddle a canoe up the same river against the current.
1. How long would it take with no current?
2. How long would it take to get down the river with the usual current without a paddle?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
May 28th, 2025 at 11:30:08 AM
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1) 35/6 = 5.83 days
2) 35 days
Last edited by: Ace2 on May 28, 2025
It’s all about making that GTA
May 28th, 2025 at 12:59:46 PM
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Let C be the speed of the canoe, and W the speed of the current
In terms of what fraction of the river is covered per day:
C + W = 1/5
C - W = 1/7
2C = 1/5 + 1/7 = 12/35 -> C = 6/35, so it would take the canoe 35/6 days, or 5 days, 20 hours
2W = 1/5 - 1/7 = 2/35 -> W = 1/35, so a drifting canoe would take 35 days
May 29th, 2025 at 3:35:55 AM
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If the river had been 35 miles then down stream with paddle would be 7mpd (miles per day). Upstream 5mpd. Hence p+x=7, p-x=5. p=6, x=1. Thus gets 35/6 and 35.
May 29th, 2025 at 7:19:55 AM
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I agree with the answers posted!
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
June 1st, 2025 at 2:21:23 AM
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Here is another classic that has probably been asked before.
There are 100 light switches, all initially in the off position.
There are 100 people, numbered 1 to 100.
Person 1 flips every switch.
Person 2 flips every 2nd switch (2, 4, 6, ...)
Person 3 flips every 3rd switch (3, 6, 9...)
Person n flips every nth switch (n, 2n, 3n -- up to 100)
After all 100 people, which switches are up?
There are 100 light switches, all initially in the off position.
There are 100 people, numbered 1 to 100.
Person 1 flips every switch.
Person 2 flips every 2nd switch (2, 4, 6, ...)
Person 3 flips every 3rd switch (3, 6, 9...)
Person n flips every nth switch (n, 2n, 3n -- up to 100)
After all 100 people, which switches are up?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
June 1st, 2025 at 8:56:08 AM
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This is usually asked with school lockers.
Variation on the problem:
There are 1000 switches, and 500 people, numbered 1, 3, 5, ..., 997, 999. How many switches will be on after these 500 people have acted?
Person N will flip every switch that is a multiple of N.
This is equivalent to, Switch N will be flipped by every person who is a factor of N.
The numbers with an odd number of factors will end up being flipped an odd number of times, and thus remain on.
Note that, pretty much by definition of factor, if K is a factor of N, then N/K is also a factor of N. Thus, every number "should" have an even number of factors.
However, for squares, if K is the square root of N, then N/K = K, so squares have an odd number of factors.
The switches that are up are 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100.
This is equivalent to, Switch N will be flipped by every person who is a factor of N.
The numbers with an odd number of factors will end up being flipped an odd number of times, and thus remain on.
Note that, pretty much by definition of factor, if K is a factor of N, then N/K is also a factor of N. Thus, every number "should" have an even number of factors.
However, for squares, if K is the square root of N, then N/K = K, so squares have an odd number of factors.
The switches that are up are 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100.
Variation on the problem:
There are 1000 switches, and 500 people, numbered 1, 3, 5, ..., 997, 999. How many switches will be on after these 500 people have acted?
June 1st, 2025 at 7:15:05 PM
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Quote: ThatDonGuy
Person N will flip every switch that is a multiple of N.
This is equivalent to, Switch N will be flipped by every person who is a factor of N.
The numbers with an odd number of factors will end up being flipped an odd number of times, and thus remain on.
Note that, pretty much by definition of factor, if K is a factor of N, then N/K is also a factor of N. Thus, every number "should" have an even number of factors.
However, for squares, if K is the square root of N, then N/K = K, so squares have an odd number of factors.
The switches that are up are 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100.
I agree!
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
June 2nd, 2025 at 3:25:17 PM
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Quote: ThatDonGuyThis is usually asked with school lockers.
Person N will flip every switch that is a multiple of N.
This is equivalent to, Switch N will be flipped by every person who is a factor of N.
The numbers with an odd number of factors will end up being flipped an odd number of times, and thus remain on.
Note that, pretty much by definition of factor, if K is a factor of N, then N/K is also a factor of N. Thus, every number "should" have an even number of factors.
However, for squares, if K is the square root of N, then N/K = K, so squares have an odd number of factors.
The switches that are up are 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100.
Variation on the problem:
There are 1000 switches, and 500 people, numbered 1, 3, 5, ..., 997, 999. How many switches will be on after these 500 people have acted?
link to original post
53
All odd squares (like the last solution) plus all 2^x (would only be flipped once by person1) plus any combination of those two. So, for instance, 392 would be on since it’s 7^2 * 2^3
All odd squares (like the last solution) plus all 2^x (would only be flipped once by person1) plus any combination of those two. So, for instance, 392 would be on since it’s 7^2 * 2^3
It’s all about making that GTA
June 2nd, 2025 at 4:15:59 PM
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Quote: Ace2Quote: ThatDonGuyThis is usually asked with school lockers.
Person N will flip every switch that is a multiple of N.
This is equivalent to, Switch N will be flipped by every person who is a factor of N.
The numbers with an odd number of factors will end up being flipped an odd number of times, and thus remain on.
Note that, pretty much by definition of factor, if K is a factor of N, then N/K is also a factor of N. Thus, every number "should" have an even number of factors.
However, for squares, if K is the square root of N, then N/K = K, so squares have an odd number of factors.
The switches that are up are 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100.
Variation on the problem:
There are 1000 switches, and 500 people, numbered 1, 3, 5, ..., 997, 999. How many switches will be on after these 500 people have acted?
link to original post53
All odd squares (like the last solution) plus all 2^x (would only be flipped once by person1) plus any combination of those two. So, for instance, 392 would be on since it’s 7^2 * 2^3
link to original post
Correct.
June 2nd, 2025 at 7:15:46 PM
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Here’s an easy variation: There are 10,000 switches, all initially in the ON position. There are 10,000 people numbered 1 to 10,000Quote: WizardHere is another classic that has probably been asked before.
There are 100 light switches, all initially in the off position.
There are 100 people, numbered 1 to 100.
Person 1 flips every switch.
Person 2 flips every 2nd switch (2, 4, 6, ...)
Person 3 flips every 3rd switch (3, 6, 9...)
Person n flips every nth switch (n, 2n, 3n -- up to 100)
After all 100 people, which switches are up?
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The switches are semi-smart…they will change position up to a maximum of three times, then they stay wherever they are
After all 10,000 people, how many switches are ON?
It’s all about making that GTA
June 2nd, 2025 at 7:35:15 PM
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Quote: Ace2Here’s an easy variation: There are 10,000 switches, all initially in the ON position. There are 10,000 people numbered 1 to 10,000
The switches are semi-smart…they will change position up to a maximum of three times, then they stay wherever they are
After all 10,000 people, how many switches are ON?
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...you know that there are 1229 primes less than 10,000
If a switch is switched three times, it will remain off. For N > 1, switch N will be switched by person 1 and by person N, so it remains on only if N has no other factors - i.e. if N is prime. Note that since 1 has only one factor, switch 1 will be switched once, and then remain off.
Thus, there will be 1229 switches that are on.
June 2nd, 2025 at 8:43:42 PM
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It’s easy if you don’t know how many primes there are <n. It’s easy because it can be answered with a couple swipes of a slide rule
Applying Ace2’s Third Conjecture, 10,000 / (Ln(10,000) - 1) gives 1,218 primes, within 1% of the exact answer. Ace2 will always accept a formulaic answer within 1% unless an exact closed-form solution is feasible
Applying Ace2’s Third Conjecture, 10,000 / (Ln(10,000) - 1) gives 1,218 primes, within 1% of the exact answer. Ace2 will always accept a formulaic answer within 1% unless an exact closed-form solution is feasible
Last edited by: Ace2 on Jun 2, 2025
It’s all about making that GTA
