Mixing
Ten people are sitting in a row, and each is thinking of a negative integer no smaller than $-15$.
$begingroup$
Ten people are sitting in a row, and each is thinking of a negative integer no smaller than $-15$. Each person subtracts, from his own number, the number of the person sitting to his right (the rightmost person does nothing). Because he has nothing else to do, the rightmost person observes that all the differences were positive. Let $x$ be the greatest integer owned by one of the 10 people at the beginning. What is the minimum possible value of $x$?
Not sure how to go about this. I think it is 1 since -14-(-15)=1. I'm not sure though.
arithmetic
asked Feb 21 '18 at 17:52
ddswsdddswsd
37917
$endgroup$
add a comment |
$begingroup$
Ten people are sitting in a row, and each is thinking of a negative integer no smaller than $-15$. Each person subtracts, from his own number, the number of the person sitting to his right (the rightmost person does nothing). Because he has nothing else to do, the rightmost person observes that all the differences were positive. Let $x$ be the greatest integer owned by one of the 10 people at the beginning. What is the minimum possible value of $x$?
Not sure how to go about this. I think it is 1 since -14-(-15)=1. I'm not sure though.
arithmetic
asked Feb 21 '18 at 17:52
ddswsdddswsd
37917
$endgroup$
$begingroup$
Yes, they are integers no smaller than -15
$endgroup$
– ddswsd
Feb 21 '18 at 18:25
add a comment |
$begingroup$
Ten people are sitting in a row, and each is thinking of a negative integer no smaller than $-15$. Each person subtracts, from his own number, the number of the person sitting to his right (the rightmost person does nothing). Because he has nothing else to do, the rightmost person observes that all the differences were positive. Let $x$ be the greatest integer owned by one of the 10 people at the beginning. What is the minimum possible value of $x$?
Not sure how to go about this. I think it is 1 since -14-(-15)=1. I'm not sure though.
arithmetic
asked Feb 21 '18 at 17:52
ddswsdddswsd
37917
$endgroup$
Ten people are sitting in a row, and each is thinking of a negative integer no smaller than $-15$. Each person subtracts, from his own number, the number of the person sitting to his right (the rightmost person does nothing). Because he has nothing else to do, the rightmost person observes that all the differences were positive. Let $x$ be the greatest integer owned by one of the 10 people at the beginning. What is the minimum possible value of $x$?
Not sure how to go about this. I think it is 1 since -14-(-15)=1. I'm not sure though.
arithmetic
arithmetic
asked Feb 21 '18 at 17:52
ddswsdddswsd
37917
asked Feb 21 '18 at 17:52
ddswsdddswsd
37917
asked Feb 21 '18 at 17:52
ddswsdddswsd
37917
asked Feb 21 '18 at 17:52
ddswsdddswsd
37917
asked Feb 21 '18 at 17:52
ddswsdddswsd
37917
37917
$begingroup$
Yes, they are integers no smaller than -15
$endgroup$
– ddswsd
Feb 21 '18 at 18:25
add a comment |
$begingroup$
Yes, they are integers no smaller than -15
$endgroup$
– ddswsd
Feb 21 '18 at 18:25
$begingroup$
Yes, they are integers no smaller than -15
$endgroup$
– ddswsd
Feb 21 '18 at 18:25
$begingroup$
Yes, they are integers no smaller than -15
$endgroup$
– ddswsd
Feb 21 '18 at 18:25
add a comment |
3 Answers
3
active
oldest
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$begingroup$
Let $x_1,x_2,dots,x_{10}$ be your numbers. Then $y_i=x_i-x_{i+1}.$ for $i=1,dots,9,$ be the results of the subtractions. Since $y_i$ are positive, then $x_igeq 1+x_{i+1}.$ In particular, $x_1$ is the largest value, and $x_1geq x_{10}+9geq -15+9=-6.$
Now you just need to find such an example where $x_1=-6.$ That is you minimum largest $x_i.$
answered Feb 21 '18 at 18:41
Thomas AndrewsThomas Andrews
130k12147298
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add a comment |
$begingroup$
As I understand it, we shall have
$-15 le x_n le -1$ and $0 < x_n -x_{n+1}$.
That is
$-15 le x_{10}<x_9< cdots < x_1 le -1$
Then greatest $x$, means $x_n le -1$ for any $n$, while "minimum possible" means that when $x_{10}=-15$ then $x_1=-6$ (?)
answered Feb 21 '18 at 18:09
G CabG Cab
19.9k31340
$endgroup$
$begingroup$
I'm not sure I understand this.
$endgroup$
– ddswsd
Feb 21 '18 at 18:25
add a comment |
$begingroup$
Each person's number has to be greater than the number of the person sitting to his right, so that when he subtracts his righthand person's number from his own, he gets a positive number. This means that the person on the far left had the largest number to start with; in other words, the leftmost person had the number $x$. This also means that the rightmost person had the smallest integer to start with.
We want to minimize $x$, so it makes sense to give the rightmost person $-15$, since that's the lowest number he could have started with. We also want to make the differences as small as possible, so let all the differences be 1 (remember, all the numbers must be integers). This means that as we go from right to left, we add 1 for each person we pass. Since there were 9 differences calculated in total (one person did nothing), we add 9 overall in going from the rightmost person to the leftmost person. Since the rightmost person had $-15$, the minimum that the leftmost person could have had is $-15+9=boxed{-6}$.
answered Jan 21 at 22:46
szaasdszaasd
1
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $x_1,x_2,dots,x_{10}$ be your numbers. Then $y_i=x_i-x_{i+1}.$ for $i=1,dots,9,$ be the results of the subtractions. Since $y_i$ are positive, then $x_igeq 1+x_{i+1}.$ In particular, $x_1$ is the largest value, and $x_1geq x_{10}+9geq -15+9=-6.$
Now you just need to find such an example where $x_1=-6.$ That is you minimum largest $x_i.$
answered Feb 21 '18 at 18:41
Thomas AndrewsThomas Andrews
130k12147298
$endgroup$
add a comment |
$begingroup$
Let $x_1,x_2,dots,x_{10}$ be your numbers. Then $y_i=x_i-x_{i+1}.$ for $i=1,dots,9,$ be the results of the subtractions. Since $y_i$ are positive, then $x_igeq 1+x_{i+1}.$ In particular, $x_1$ is the largest value, and $x_1geq x_{10}+9geq -15+9=-6.$
Now you just need to find such an example where $x_1=-6.$ That is you minimum largest $x_i.$
answered Feb 21 '18 at 18:41
Thomas AndrewsThomas Andrews
130k12147298
$endgroup$
add a comment |
$begingroup$
Let $x_1,x_2,dots,x_{10}$ be your numbers. Then $y_i=x_i-x_{i+1}.$ for $i=1,dots,9,$ be the results of the subtractions. Since $y_i$ are positive, then $x_igeq 1+x_{i+1}.$ In particular, $x_1$ is the largest value, and $x_1geq x_{10}+9geq -15+9=-6.$
Now you just need to find such an example where $x_1=-6.$ That is you minimum largest $x_i.$
answered Feb 21 '18 at 18:41
Thomas AndrewsThomas Andrews
130k12147298
$endgroup$
Let $x_1,x_2,dots,x_{10}$ be your numbers. Then $y_i=x_i-x_{i+1}.$ for $i=1,dots,9,$ be the results of the subtractions. Since $y_i$ are positive, then $x_igeq 1+x_{i+1}.$ In particular, $x_1$ is the largest value, and $x_1geq x_{10}+9geq -15+9=-6.$
Now you just need to find such an example where $x_1=-6.$ That is you minimum largest $x_i.$
answered Feb 21 '18 at 18:41
Thomas AndrewsThomas Andrews
130k12147298
answered Feb 21 '18 at 18:41
Thomas AndrewsThomas Andrews
130k12147298
answered Feb 21 '18 at 18:41
Thomas AndrewsThomas Andrews
130k12147298
answered Feb 21 '18 at 18:41
Thomas AndrewsThomas Andrews
130k12147298
130k12147298
add a comment |
add a comment |
$begingroup$
As I understand it, we shall have
$-15 le x_n le -1$ and $0 < x_n -x_{n+1}$.
That is
$-15 le x_{10}<x_9< cdots < x_1 le -1$
Then greatest $x$, means $x_n le -1$ for any $n$, while "minimum possible" means that when $x_{10}=-15$ then $x_1=-6$ (?)
answered Feb 21 '18 at 18:09
G CabG Cab
19.9k31340
$endgroup$
$begingroup$
I'm not sure I understand this.
$endgroup$
– ddswsd
Feb 21 '18 at 18:25
add a comment |
$begingroup$
As I understand it, we shall have
$-15 le x_n le -1$ and $0 < x_n -x_{n+1}$.
That is
$-15 le x_{10}<x_9< cdots < x_1 le -1$
Then greatest $x$, means $x_n le -1$ for any $n$, while "minimum possible" means that when $x_{10}=-15$ then $x_1=-6$ (?)
answered Feb 21 '18 at 18:09
G CabG Cab
19.9k31340
$endgroup$
$begingroup$
I'm not sure I understand this.
$endgroup$
– ddswsd
Feb 21 '18 at 18:25
add a comment |
$begingroup$
As I understand it, we shall have
$-15 le x_n le -1$ and $0 < x_n -x_{n+1}$.
That is
$-15 le x_{10}<x_9< cdots < x_1 le -1$
Then greatest $x$, means $x_n le -1$ for any $n$, while "minimum possible" means that when $x_{10}=-15$ then $x_1=-6$ (?)
answered Feb 21 '18 at 18:09
G CabG Cab
19.9k31340
$endgroup$
As I understand it, we shall have
$-15 le x_n le -1$ and $0 < x_n -x_{n+1}$.
That is
$-15 le x_{10}<x_9< cdots < x_1 le -1$
Then greatest $x$, means $x_n le -1$ for any $n$, while "minimum possible" means that when $x_{10}=-15$ then $x_1=-6$ (?)
answered Feb 21 '18 at 18:09
G CabG Cab
19.9k31340
edited Feb 21 '18 at 18:55
answered Feb 21 '18 at 18:09
G CabG Cab
19.9k31340
answered Feb 21 '18 at 18:09
G CabG Cab
19.9k31340
answered Feb 21 '18 at 18:09
G CabG Cab
19.9k31340
19.9k31340
$begingroup$
I'm not sure I understand this.
$endgroup$
– ddswsd
Feb 21 '18 at 18:25
add a comment |
$begingroup$
I'm not sure I understand this.
$endgroup$
– ddswsd
Feb 21 '18 at 18:25
$begingroup$
I'm not sure I understand this.
$endgroup$
– ddswsd
Feb 21 '18 at 18:25
$begingroup$
I'm not sure I understand this.
$endgroup$
– ddswsd
Feb 21 '18 at 18:25
add a comment |
$begingroup$
Each person's number has to be greater than the number of the person sitting to his right, so that when he subtracts his righthand person's number from his own, he gets a positive number. This means that the person on the far left had the largest number to start with; in other words, the leftmost person had the number $x$. This also means that the rightmost person had the smallest integer to start with.
We want to minimize $x$, so it makes sense to give the rightmost person $-15$, since that's the lowest number he could have started with. We also want to make the differences as small as possible, so let all the differences be 1 (remember, all the numbers must be integers). This means that as we go from right to left, we add 1 for each person we pass. Since there were 9 differences calculated in total (one person did nothing), we add 9 overall in going from the rightmost person to the leftmost person. Since the rightmost person had $-15$, the minimum that the leftmost person could have had is $-15+9=boxed{-6}$.
answered Jan 21 at 22:46
szaasdszaasd
1
$endgroup$
add a comment |
$begingroup$
Each person's number has to be greater than the number of the person sitting to his right, so that when he subtracts his righthand person's number from his own, he gets a positive number. This means that the person on the far left had the largest number to start with; in other words, the leftmost person had the number $x$. This also means that the rightmost person had the smallest integer to start with.
We want to minimize $x$, so it makes sense to give the rightmost person $-15$, since that's the lowest number he could have started with. We also want to make the differences as small as possible, so let all the differences be 1 (remember, all the numbers must be integers). This means that as we go from right to left, we add 1 for each person we pass. Since there were 9 differences calculated in total (one person did nothing), we add 9 overall in going from the rightmost person to the leftmost person. Since the rightmost person had $-15$, the minimum that the leftmost person could have had is $-15+9=boxed{-6}$.
answered Jan 21 at 22:46
szaasdszaasd
1
$endgroup$
add a comment |
$begingroup$
Each person's number has to be greater than the number of the person sitting to his right, so that when he subtracts his righthand person's number from his own, he gets a positive number. This means that the person on the far left had the largest number to start with; in other words, the leftmost person had the number $x$. This also means that the rightmost person had the smallest integer to start with.
We want to minimize $x$, so it makes sense to give the rightmost person $-15$, since that's the lowest number he could have started with. We also want to make the differences as small as possible, so let all the differences be 1 (remember, all the numbers must be integers). This means that as we go from right to left, we add 1 for each person we pass. Since there were 9 differences calculated in total (one person did nothing), we add 9 overall in going from the rightmost person to the leftmost person. Since the rightmost person had $-15$, the minimum that the leftmost person could have had is $-15+9=boxed{-6}$.
answered Jan 21 at 22:46
szaasdszaasd
1
$endgroup$
Each person's number has to be greater than the number of the person sitting to his right, so that when he subtracts his righthand person's number from his own, he gets a positive number. This means that the person on the far left had the largest number to start with; in other words, the leftmost person had the number $x$. This also means that the rightmost person had the smallest integer to start with.
We want to minimize $x$, so it makes sense to give the rightmost person $-15$, since that's the lowest number he could have started with. We also want to make the differences as small as possible, so let all the differences be 1 (remember, all the numbers must be integers). This means that as we go from right to left, we add 1 for each person we pass. Since there were 9 differences calculated in total (one person did nothing), we add 9 overall in going from the rightmost person to the leftmost person. Since the rightmost person had $-15$, the minimum that the leftmost person could have had is $-15+9=boxed{-6}$.
answered Jan 21 at 22:46
szaasdszaasd
1
answered Jan 21 at 22:46
szaasdszaasd
1
answered Jan 21 at 22:46
szaasdszaasd
1
answered Jan 21 at 22:46
szaasdszaasd
1
1
add a comment |
add a comment |
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$begingroup$
Yes, they are integers no smaller than -15
$endgroup$
– ddswsd
Feb 21 '18 at 18:25