Different Differences between an Arrangement of NumbersA list of numbers andAverage of NumbersPossibilities with unit digits and numbersFind two numbers, given their difference and quotientArrangement of Necklace beads and membersA petroleum company has two different sources of crude oil.Why does the difference between 150% and 120% equal 25%?Consecutive Square NumbersDistance between two foci - astronomyWhat numbers should I not pick?
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Different Differences between an Arrangement of Numbers
A list of numbers andAverage of NumbersPossibilities with unit digits and numbersFind two numbers, given their difference and quotientArrangement of Necklace beads and membersA petroleum company has two different sources of crude oil.Why does the difference between 150% and 120% equal 25%?Consecutive Square NumbersDistance between two foci - astronomyWhat numbers should I not pick?
$begingroup$
This was an optional problem given to me which I just don't know how to approach.
The full question states: "An arrangement of numbers has 'different differences' when the differences between all the neighbouring numbers are different-
e.g. '1423' the differences between these neighbours (in order) are 3, 2, 1
Given numbers 1 - 6 and the 3rd number being 3 (can't use it again), what is the sum of the last three numbers?"
basically in the format: _ _ 3 _ _ _
I initially looked at the example and tried to find any clues, e.g descending differences (/maybe ascending differences) and relationships between the actual neighbouring number, quickly found that this wasn't the right way to do this. Then just decided to play around and try some numbers that I thought would work, didn't see a new perspective or similar and kinda just pondered on until the next problem. However, I still really want any insight/a hint or something to lead me in the right direction to be able to solve this. Any help is welcome, Thanks.
word-problem
asked Mar 21 at 6:09
thethe
224
$endgroup$
add a comment |
$begingroup$
This was an optional problem given to me which I just don't know how to approach.
The full question states: "An arrangement of numbers has 'different differences' when the differences between all the neighbouring numbers are different-
e.g. '1423' the differences between these neighbours (in order) are 3, 2, 1
Given numbers 1 - 6 and the 3rd number being 3 (can't use it again), what is the sum of the last three numbers?"
basically in the format: _ _ 3 _ _ _
I initially looked at the example and tried to find any clues, e.g descending differences (/maybe ascending differences) and relationships between the actual neighbouring number, quickly found that this wasn't the right way to do this. Then just decided to play around and try some numbers that I thought would work, didn't see a new perspective or similar and kinda just pondered on until the next problem. However, I still really want any insight/a hint or something to lead me in the right direction to be able to solve this. Any help is welcome, Thanks.
word-problem
asked Mar 21 at 6:09
thethe
224
$endgroup$
$begingroup$
Unless someone has a surprising answer, I can't think of a means to solve this that wouldn't be at some level brute-force/guessing. I also wonder if there definitely are not multiple solutions to this - would be interesting if there were multiple solutions yet all three had the same sum...
$endgroup$
– Eevee Trainer
Mar 21 at 6:15
1
$begingroup$
This was taken from a multiple choice paper, I could technically guess the answer (the answers if you want are: 12, 13, 14, 15,16), but I want to learn how I would solve this, say if it was not on a multiple choice paper. If this extra info helps you though, please do use it.
$endgroup$
– the
Mar 21 at 6:18
$begingroup$
On just screwing around a bit, I did find one valid sequence - $4,2,3,6,1,5$ with the differences $2,1,3,5,4$. The sum in this case would be $12$. That said I wasn't really following any particular pattern or anything, so I doubt it'd be of much use. On further note, I wonder if we can eliminate any of the answer choices: for example, $16$ can be eliminated (if the end has $4,5,6$ - not necessarily in that order - the sum is $15$ at most).
$endgroup$
– Eevee Trainer
Mar 21 at 6:28
$begingroup$
Oh, seems I should've messed around even more! But yeah as you said I could have done the process of elimination or just went through every answer and exhaust the possibilities. Maybe they made 12 the first option, on purpose. Anyways thank you for the answer, I'm going to now see what happens if I do 5 numbers instead.
$endgroup$
– the
Mar 21 at 6:36
add a comment |
$begingroup$
This was an optional problem given to me which I just don't know how to approach.
The full question states: "An arrangement of numbers has 'different differences' when the differences between all the neighbouring numbers are different-
e.g. '1423' the differences between these neighbours (in order) are 3, 2, 1
Given numbers 1 - 6 and the 3rd number being 3 (can't use it again), what is the sum of the last three numbers?"
basically in the format: _ _ 3 _ _ _
I initially looked at the example and tried to find any clues, e.g descending differences (/maybe ascending differences) and relationships between the actual neighbouring number, quickly found that this wasn't the right way to do this. Then just decided to play around and try some numbers that I thought would work, didn't see a new perspective or similar and kinda just pondered on until the next problem. However, I still really want any insight/a hint or something to lead me in the right direction to be able to solve this. Any help is welcome, Thanks.
word-problem
asked Mar 21 at 6:09
thethe
224
$endgroup$
This was an optional problem given to me which I just don't know how to approach.
The full question states: "An arrangement of numbers has 'different differences' when the differences between all the neighbouring numbers are different-
e.g. '1423' the differences between these neighbours (in order) are 3, 2, 1
Given numbers 1 - 6 and the 3rd number being 3 (can't use it again), what is the sum of the last three numbers?"
basically in the format: _ _ 3 _ _ _
I initially looked at the example and tried to find any clues, e.g descending differences (/maybe ascending differences) and relationships between the actual neighbouring number, quickly found that this wasn't the right way to do this. Then just decided to play around and try some numbers that I thought would work, didn't see a new perspective or similar and kinda just pondered on until the next problem. However, I still really want any insight/a hint or something to lead me in the right direction to be able to solve this. Any help is welcome, Thanks.
word-problem
word-problem
asked Mar 21 at 6:09
thethe
224
asked Mar 21 at 6:09
thethe
224
asked Mar 21 at 6:09
thethe
224
asked Mar 21 at 6:09
thethe
224
asked Mar 21 at 6:09
thethe
224
224
$begingroup$
Unless someone has a surprising answer, I can't think of a means to solve this that wouldn't be at some level brute-force/guessing. I also wonder if there definitely are not multiple solutions to this - would be interesting if there were multiple solutions yet all three had the same sum...
$endgroup$
– Eevee Trainer
Mar 21 at 6:15
1
$begingroup$
This was taken from a multiple choice paper, I could technically guess the answer (the answers if you want are: 12, 13, 14, 15,16), but I want to learn how I would solve this, say if it was not on a multiple choice paper. If this extra info helps you though, please do use it.
$endgroup$
– the
Mar 21 at 6:18
$begingroup$
On just screwing around a bit, I did find one valid sequence - $4,2,3,6,1,5$ with the differences $2,1,3,5,4$. The sum in this case would be $12$. That said I wasn't really following any particular pattern or anything, so I doubt it'd be of much use. On further note, I wonder if we can eliminate any of the answer choices: for example, $16$ can be eliminated (if the end has $4,5,6$ - not necessarily in that order - the sum is $15$ at most).
$endgroup$
– Eevee Trainer
Mar 21 at 6:28
$begingroup$
Oh, seems I should've messed around even more! But yeah as you said I could have done the process of elimination or just went through every answer and exhaust the possibilities. Maybe they made 12 the first option, on purpose. Anyways thank you for the answer, I'm going to now see what happens if I do 5 numbers instead.
$endgroup$
– the
Mar 21 at 6:36
add a comment |
$begingroup$
Unless someone has a surprising answer, I can't think of a means to solve this that wouldn't be at some level brute-force/guessing. I also wonder if there definitely are not multiple solutions to this - would be interesting if there were multiple solutions yet all three had the same sum...
$endgroup$
– Eevee Trainer
Mar 21 at 6:15
1
$begingroup$
This was taken from a multiple choice paper, I could technically guess the answer (the answers if you want are: 12, 13, 14, 15,16), but I want to learn how I would solve this, say if it was not on a multiple choice paper. If this extra info helps you though, please do use it.
$endgroup$
– the
Mar 21 at 6:18
$begingroup$
On just screwing around a bit, I did find one valid sequence - $4,2,3,6,1,5$ with the differences $2,1,3,5,4$. The sum in this case would be $12$. That said I wasn't really following any particular pattern or anything, so I doubt it'd be of much use. On further note, I wonder if we can eliminate any of the answer choices: for example, $16$ can be eliminated (if the end has $4,5,6$ - not necessarily in that order - the sum is $15$ at most).
$endgroup$
– Eevee Trainer
Mar 21 at 6:28
$begingroup$
Oh, seems I should've messed around even more! But yeah as you said I could have done the process of elimination or just went through every answer and exhaust the possibilities. Maybe they made 12 the first option, on purpose. Anyways thank you for the answer, I'm going to now see what happens if I do 5 numbers instead.
$endgroup$
– the
Mar 21 at 6:36
$begingroup$
Unless someone has a surprising answer, I can't think of a means to solve this that wouldn't be at some level brute-force/guessing. I also wonder if there definitely are not multiple solutions to this - would be interesting if there were multiple solutions yet all three had the same sum...
$endgroup$
– Eevee Trainer
Mar 21 at 6:15
$begingroup$
Unless someone has a surprising answer, I can't think of a means to solve this that wouldn't be at some level brute-force/guessing. I also wonder if there definitely are not multiple solutions to this - would be interesting if there were multiple solutions yet all three had the same sum...
$endgroup$
– Eevee Trainer
Mar 21 at 6:15
1
1
$begingroup$
This was taken from a multiple choice paper, I could technically guess the answer (the answers if you want are: 12, 13, 14, 15,16), but I want to learn how I would solve this, say if it was not on a multiple choice paper. If this extra info helps you though, please do use it.
$endgroup$
– the
Mar 21 at 6:18
$begingroup$
This was taken from a multiple choice paper, I could technically guess the answer (the answers if you want are: 12, 13, 14, 15,16), but I want to learn how I would solve this, say if it was not on a multiple choice paper. If this extra info helps you though, please do use it.
$endgroup$
– the
Mar 21 at 6:18
$begingroup$
On just screwing around a bit, I did find one valid sequence - $4,2,3,6,1,5$ with the differences $2,1,3,5,4$. The sum in this case would be $12$. That said I wasn't really following any particular pattern or anything, so I doubt it'd be of much use. On further note, I wonder if we can eliminate any of the answer choices: for example, $16$ can be eliminated (if the end has $4,5,6$ - not necessarily in that order - the sum is $15$ at most).
$endgroup$
– Eevee Trainer
Mar 21 at 6:28
$begingroup$
On just screwing around a bit, I did find one valid sequence - $4,2,3,6,1,5$ with the differences $2,1,3,5,4$. The sum in this case would be $12$. That said I wasn't really following any particular pattern or anything, so I doubt it'd be of much use. On further note, I wonder if we can eliminate any of the answer choices: for example, $16$ can be eliminated (if the end has $4,5,6$ - not necessarily in that order - the sum is $15$ at most).
$endgroup$
– Eevee Trainer
Mar 21 at 6:28
$begingroup$
Oh, seems I should've messed around even more! But yeah as you said I could have done the process of elimination or just went through every answer and exhaust the possibilities. Maybe they made 12 the first option, on purpose. Anyways thank you for the answer, I'm going to now see what happens if I do 5 numbers instead.
$endgroup$
– the
Mar 21 at 6:36
$begingroup$
Oh, seems I should've messed around even more! But yeah as you said I could have done the process of elimination or just went through every answer and exhaust the possibilities. Maybe they made 12 the first option, on purpose. Anyways thank you for the answer, I'm going to now see what happens if I do 5 numbers instead.
$endgroup$
– the
Mar 21 at 6:36
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
[This solution uses the fact, mentioned in a comment, that the sum of the last three numbers is one of $12, 13, 14, 15$ or $16$.]
The possible differences are $1,2,3,4,5$ and each must occur. The only way to produce a difference of $5$ is for $1$ and $6$ to be adjacent. The difference of $4$ requires either $1$ adjacent to $5$ or $2$ adjacent to $6$, neither of which is possible if $1$ and $6$ are placed before the $3$. Therefore they are after, and the maximum possible sum of the last three numbers is $1+6+5=12$. Since the other choices are even larger, $12$ must be correct.
answered Mar 21 at 8:57
FredHFredH
3,6501023
$endgroup$
add a comment |
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$begingroup$
[This solution uses the fact, mentioned in a comment, that the sum of the last three numbers is one of $12, 13, 14, 15$ or $16$.]
The possible differences are $1,2,3,4,5$ and each must occur. The only way to produce a difference of $5$ is for $1$ and $6$ to be adjacent. The difference of $4$ requires either $1$ adjacent to $5$ or $2$ adjacent to $6$, neither of which is possible if $1$ and $6$ are placed before the $3$. Therefore they are after, and the maximum possible sum of the last three numbers is $1+6+5=12$. Since the other choices are even larger, $12$ must be correct.
answered Mar 21 at 8:57
FredHFredH
3,6501023
$endgroup$
add a comment |
$begingroup$
[This solution uses the fact, mentioned in a comment, that the sum of the last three numbers is one of $12, 13, 14, 15$ or $16$.]
The possible differences are $1,2,3,4,5$ and each must occur. The only way to produce a difference of $5$ is for $1$ and $6$ to be adjacent. The difference of $4$ requires either $1$ adjacent to $5$ or $2$ adjacent to $6$, neither of which is possible if $1$ and $6$ are placed before the $3$. Therefore they are after, and the maximum possible sum of the last three numbers is $1+6+5=12$. Since the other choices are even larger, $12$ must be correct.
answered Mar 21 at 8:57
FredHFredH
3,6501023
$endgroup$
add a comment |
$begingroup$
[This solution uses the fact, mentioned in a comment, that the sum of the last three numbers is one of $12, 13, 14, 15$ or $16$.]
The possible differences are $1,2,3,4,5$ and each must occur. The only way to produce a difference of $5$ is for $1$ and $6$ to be adjacent. The difference of $4$ requires either $1$ adjacent to $5$ or $2$ adjacent to $6$, neither of which is possible if $1$ and $6$ are placed before the $3$. Therefore they are after, and the maximum possible sum of the last three numbers is $1+6+5=12$. Since the other choices are even larger, $12$ must be correct.
answered Mar 21 at 8:57
FredHFredH
3,6501023
$endgroup$
[This solution uses the fact, mentioned in a comment, that the sum of the last three numbers is one of $12, 13, 14, 15$ or $16$.]
The possible differences are $1,2,3,4,5$ and each must occur. The only way to produce a difference of $5$ is for $1$ and $6$ to be adjacent. The difference of $4$ requires either $1$ adjacent to $5$ or $2$ adjacent to $6$, neither of which is possible if $1$ and $6$ are placed before the $3$. Therefore they are after, and the maximum possible sum of the last three numbers is $1+6+5=12$. Since the other choices are even larger, $12$ must be correct.
answered Mar 21 at 8:57
FredHFredH
3,6501023
answered Mar 21 at 8:57
FredHFredH
3,6501023
answered Mar 21 at 8:57
FredHFredH
3,6501023
answered Mar 21 at 8:57
FredHFredH
3,6501023
3,6501023
add a comment |
add a comment |
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$begingroup$
Unless someone has a surprising answer, I can't think of a means to solve this that wouldn't be at some level brute-force/guessing. I also wonder if there definitely are not multiple solutions to this - would be interesting if there were multiple solutions yet all three had the same sum...
$endgroup$
– Eevee Trainer
Mar 21 at 6:15
1
$begingroup$
This was taken from a multiple choice paper, I could technically guess the answer (the answers if you want are: 12, 13, 14, 15,16), but I want to learn how I would solve this, say if it was not on a multiple choice paper. If this extra info helps you though, please do use it.
$endgroup$
– the
Mar 21 at 6:18
$begingroup$
On just screwing around a bit, I did find one valid sequence - $4,2,3,6,1,5$ with the differences $2,1,3,5,4$. The sum in this case would be $12$. That said I wasn't really following any particular pattern or anything, so I doubt it'd be of much use. On further note, I wonder if we can eliminate any of the answer choices: for example, $16$ can be eliminated (if the end has $4,5,6$ - not necessarily in that order - the sum is $15$ at most).
$endgroup$
– Eevee Trainer
Mar 21 at 6:28
$begingroup$
Oh, seems I should've messed around even more! But yeah as you said I could have done the process of elimination or just went through every answer and exhaust the possibilities. Maybe they made 12 the first option, on purpose. Anyways thank you for the answer, I'm going to now see what happens if I do 5 numbers instead.
$endgroup$
– the
Mar 21 at 6:36