Mixing
A question from my high school Olympiad.
$begingroup$
A set contains four numbers. The six pairwise sums of distinct elements of the set, in no particular order, are $189, 320, 287, 264, x, y$. Find the greatest possible value of: $x + y$.
By assuming the elements to be $a, b, c, d$ and by making $6$ different equations, I solved for $x+y$. The result came to be $793$. I am not sure if this is correct or not. Can anyone help me out?
elementary-number-theory contest-math
edited Jan 25 at 14:15
Jyrki Lahtonen
110k13171386
asked Jan 25 at 13:53
S.NepS.Nep
776
$endgroup$
add a comment |
$begingroup$
A set contains four numbers. The six pairwise sums of distinct elements of the set, in no particular order, are $189, 320, 287, 264, x, y$. Find the greatest possible value of: $x + y$.
By assuming the elements to be $a, b, c, d$ and by making $6$ different equations, I solved for $x+y$. The result came to be $793$. I am not sure if this is correct or not. Can anyone help me out?
elementary-number-theory contest-math
edited Jan 25 at 14:15
Jyrki Lahtonen
110k13171386
asked Jan 25 at 13:53
S.NepS.Nep
776
$endgroup$
$begingroup$
Sounds good to me
$endgroup$
– lioness99a
Jan 25 at 14:11
$begingroup$
@lioness99a so it's correct right?
$endgroup$
– S.Nep
Jan 25 at 14:12
$begingroup$
Without working through the equations myself I can't be 100% sure but the method sounds alright
$endgroup$
– lioness99a
Jan 25 at 14:19
$begingroup$
Are you sure it is 793, I get 761 as the maximum.
$endgroup$
– Satish Ramanathan
Jan 25 at 14:41
1
$begingroup$
Agree with Satish, I'm getting 761 as well.
$endgroup$
– Dubs
Jan 25 at 14:46
add a comment |
$begingroup$
A set contains four numbers. The six pairwise sums of distinct elements of the set, in no particular order, are $189, 320, 287, 264, x, y$. Find the greatest possible value of: $x + y$.
By assuming the elements to be $a, b, c, d$ and by making $6$ different equations, I solved for $x+y$. The result came to be $793$. I am not sure if this is correct or not. Can anyone help me out?
elementary-number-theory contest-math
edited Jan 25 at 14:15
Jyrki Lahtonen
110k13171386
asked Jan 25 at 13:53
S.NepS.Nep
776
$endgroup$
A set contains four numbers. The six pairwise sums of distinct elements of the set, in no particular order, are $189, 320, 287, 264, x, y$. Find the greatest possible value of: $x + y$.
By assuming the elements to be $a, b, c, d$ and by making $6$ different equations, I solved for $x+y$. The result came to be $793$. I am not sure if this is correct or not. Can anyone help me out?
elementary-number-theory contest-math
elementary-number-theory contest-math
edited Jan 25 at 14:15
Jyrki Lahtonen
110k13171386
asked Jan 25 at 13:53
S.NepS.Nep
776
edited Jan 25 at 14:15
Jyrki Lahtonen
110k13171386
asked Jan 25 at 13:53
S.NepS.Nep
776
edited Jan 25 at 14:15
Jyrki Lahtonen
110k13171386
edited Jan 25 at 14:15
Jyrki Lahtonen
110k13171386
edited Jan 25 at 14:15
Jyrki Lahtonen
110k13171386
110k13171386
asked Jan 25 at 13:53
S.NepS.Nep
776
asked Jan 25 at 13:53
S.NepS.Nep
776
asked Jan 25 at 13:53
S.NepS.Nep
776
776
$begingroup$
Sounds good to me
$endgroup$
– lioness99a
Jan 25 at 14:11
$begingroup$
@lioness99a so it's correct right?
$endgroup$
– S.Nep
Jan 25 at 14:12
$begingroup$
Without working through the equations myself I can't be 100% sure but the method sounds alright
$endgroup$
– lioness99a
Jan 25 at 14:19
$begingroup$
Are you sure it is 793, I get 761 as the maximum.
$endgroup$
– Satish Ramanathan
Jan 25 at 14:41
1
$begingroup$
Agree with Satish, I'm getting 761 as well.
$endgroup$
– Dubs
Jan 25 at 14:46
add a comment |
$begingroup$
Sounds good to me
$endgroup$
– lioness99a
Jan 25 at 14:11
$begingroup$
@lioness99a so it's correct right?
$endgroup$
– S.Nep
Jan 25 at 14:12
$begingroup$
Without working through the equations myself I can't be 100% sure but the method sounds alright
$endgroup$
– lioness99a
Jan 25 at 14:19
$begingroup$
Are you sure it is 793, I get 761 as the maximum.
$endgroup$
– Satish Ramanathan
Jan 25 at 14:41
1
$begingroup$
Agree with Satish, I'm getting 761 as well.
$endgroup$
– Dubs
Jan 25 at 14:46
$begingroup$
Sounds good to me
$endgroup$
– lioness99a
Jan 25 at 14:11
$begingroup$
Sounds good to me
$endgroup$
– lioness99a
Jan 25 at 14:11
$begingroup$
@lioness99a so it's correct right?
$endgroup$
– S.Nep
Jan 25 at 14:12
$begingroup$
@lioness99a so it's correct right?
$endgroup$
– S.Nep
Jan 25 at 14:12
$begingroup$
Without working through the equations myself I can't be 100% sure but the method sounds alright
$endgroup$
– lioness99a
Jan 25 at 14:19
$begingroup$
Without working through the equations myself I can't be 100% sure but the method sounds alright
$endgroup$
– lioness99a
Jan 25 at 14:19
$begingroup$
Are you sure it is 793, I get 761 as the maximum.
$endgroup$
– Satish Ramanathan
Jan 25 at 14:41
$begingroup$
Are you sure it is 793, I get 761 as the maximum.
$endgroup$
– Satish Ramanathan
Jan 25 at 14:41
1
1
$begingroup$
Agree with Satish, I'm getting 761 as well.
$endgroup$
– Dubs
Jan 25 at 14:46
$begingroup$
Agree with Satish, I'm getting 761 as well.
$endgroup$
– Dubs
Jan 25 at 14:46
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Taking the hint from @Dubs, you want d to be the largest, b and c the next larger than a. For this to happen, you get ($d+a = 320, b+c = 287, b+a = 264 and c+a = 189$).
From the last two weg get $b-c = 75$ and from this and the second one we get $2b = 362$, giving you $b = 181, a =83, c = 106$ and $d= 237$ and hence $x+y = 761$
answered Jan 25 at 14:48
Satish RamanathanSatish Ramanathan
10k31323
$endgroup$
$begingroup$
I believe that $a+d$ or $b+c$ can be interchanged as the 3rd/4th smallest values. But regardless, they will arrive to 761.
$endgroup$
– Dubs
Jan 25 at 15:04
$begingroup$
Ah. Yes. Got it. Thanks.
$endgroup$
– S.Nep
Jan 25 at 15:56
add a comment |
$begingroup$
You are off to a good start. But you want a stronger assumption. Otherwise, you will have a lot of cases to consider. For example, if $a+b=189$ should $a+c = 264$ or $c+d = 264$?
Try to assume that $a le b le c le d$ instead.
This will still run into some choices. But those choices won't matter if you focus on solving for maximum value of $x + y$ so don't actually need to calculate $a,b,c,d$ individually.
Additional hint:
Can you see why to get the maximum value of $x + y$ then $x + y = b + c + 2d$?
answered Jan 25 at 14:37
DubsDubs
55926
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Taking the hint from @Dubs, you want d to be the largest, b and c the next larger than a. For this to happen, you get ($d+a = 320, b+c = 287, b+a = 264 and c+a = 189$).
From the last two weg get $b-c = 75$ and from this and the second one we get $2b = 362$, giving you $b = 181, a =83, c = 106$ and $d= 237$ and hence $x+y = 761$
answered Jan 25 at 14:48
Satish RamanathanSatish Ramanathan
10k31323
$endgroup$
$begingroup$
I believe that $a+d$ or $b+c$ can be interchanged as the 3rd/4th smallest values. But regardless, they will arrive to 761.
$endgroup$
– Dubs
Jan 25 at 15:04
$begingroup$
Ah. Yes. Got it. Thanks.
$endgroup$
– S.Nep
Jan 25 at 15:56
add a comment |
$begingroup$
Taking the hint from @Dubs, you want d to be the largest, b and c the next larger than a. For this to happen, you get ($d+a = 320, b+c = 287, b+a = 264 and c+a = 189$).
From the last two weg get $b-c = 75$ and from this and the second one we get $2b = 362$, giving you $b = 181, a =83, c = 106$ and $d= 237$ and hence $x+y = 761$
answered Jan 25 at 14:48
Satish RamanathanSatish Ramanathan
10k31323
$endgroup$
$begingroup$
I believe that $a+d$ or $b+c$ can be interchanged as the 3rd/4th smallest values. But regardless, they will arrive to 761.
$endgroup$
– Dubs
Jan 25 at 15:04
$begingroup$
Ah. Yes. Got it. Thanks.
$endgroup$
– S.Nep
Jan 25 at 15:56
add a comment |
$begingroup$
Taking the hint from @Dubs, you want d to be the largest, b and c the next larger than a. For this to happen, you get ($d+a = 320, b+c = 287, b+a = 264 and c+a = 189$).
From the last two weg get $b-c = 75$ and from this and the second one we get $2b = 362$, giving you $b = 181, a =83, c = 106$ and $d= 237$ and hence $x+y = 761$
answered Jan 25 at 14:48
Satish RamanathanSatish Ramanathan
10k31323
$endgroup$
Taking the hint from @Dubs, you want d to be the largest, b and c the next larger than a. For this to happen, you get ($d+a = 320, b+c = 287, b+a = 264 and c+a = 189$).
From the last two weg get $b-c = 75$ and from this and the second one we get $2b = 362$, giving you $b = 181, a =83, c = 106$ and $d= 237$ and hence $x+y = 761$
answered Jan 25 at 14:48
Satish RamanathanSatish Ramanathan
10k31323
answered Jan 25 at 14:48
Satish RamanathanSatish Ramanathan
10k31323
answered Jan 25 at 14:48
Satish RamanathanSatish Ramanathan
10k31323
answered Jan 25 at 14:48
Satish RamanathanSatish Ramanathan
10k31323
10k31323
$begingroup$
I believe that $a+d$ or $b+c$ can be interchanged as the 3rd/4th smallest values. But regardless, they will arrive to 761.
$endgroup$
– Dubs
Jan 25 at 15:04
$begingroup$
Ah. Yes. Got it. Thanks.
$endgroup$
– S.Nep
Jan 25 at 15:56
add a comment |
$begingroup$
I believe that $a+d$ or $b+c$ can be interchanged as the 3rd/4th smallest values. But regardless, they will arrive to 761.
$endgroup$
– Dubs
Jan 25 at 15:04
$begingroup$
Ah. Yes. Got it. Thanks.
$endgroup$
– S.Nep
Jan 25 at 15:56
$begingroup$
I believe that $a+d$ or $b+c$ can be interchanged as the 3rd/4th smallest values. But regardless, they will arrive to 761.
$endgroup$
– Dubs
Jan 25 at 15:04
$begingroup$
I believe that $a+d$ or $b+c$ can be interchanged as the 3rd/4th smallest values. But regardless, they will arrive to 761.
$endgroup$
– Dubs
Jan 25 at 15:04
$begingroup$
Ah. Yes. Got it. Thanks.
$endgroup$
– S.Nep
Jan 25 at 15:56
$begingroup$
Ah. Yes. Got it. Thanks.
$endgroup$
– S.Nep
Jan 25 at 15:56
add a comment |
$begingroup$
You are off to a good start. But you want a stronger assumption. Otherwise, you will have a lot of cases to consider. For example, if $a+b=189$ should $a+c = 264$ or $c+d = 264$?
Try to assume that $a le b le c le d$ instead.
This will still run into some choices. But those choices won't matter if you focus on solving for maximum value of $x + y$ so don't actually need to calculate $a,b,c,d$ individually.
Additional hint:
Can you see why to get the maximum value of $x + y$ then $x + y = b + c + 2d$?
answered Jan 25 at 14:37
DubsDubs
55926
$endgroup$
add a comment |
$begingroup$
You are off to a good start. But you want a stronger assumption. Otherwise, you will have a lot of cases to consider. For example, if $a+b=189$ should $a+c = 264$ or $c+d = 264$?
Try to assume that $a le b le c le d$ instead.
This will still run into some choices. But those choices won't matter if you focus on solving for maximum value of $x + y$ so don't actually need to calculate $a,b,c,d$ individually.
Additional hint:
Can you see why to get the maximum value of $x + y$ then $x + y = b + c + 2d$?
answered Jan 25 at 14:37
DubsDubs
55926
$endgroup$
add a comment |
$begingroup$
You are off to a good start. But you want a stronger assumption. Otherwise, you will have a lot of cases to consider. For example, if $a+b=189$ should $a+c = 264$ or $c+d = 264$?
Try to assume that $a le b le c le d$ instead.
This will still run into some choices. But those choices won't matter if you focus on solving for maximum value of $x + y$ so don't actually need to calculate $a,b,c,d$ individually.
Additional hint:
Can you see why to get the maximum value of $x + y$ then $x + y = b + c + 2d$?
answered Jan 25 at 14:37
DubsDubs
55926
$endgroup$
You are off to a good start. But you want a stronger assumption. Otherwise, you will have a lot of cases to consider. For example, if $a+b=189$ should $a+c = 264$ or $c+d = 264$?
Try to assume that $a le b le c le d$ instead.
This will still run into some choices. But those choices won't matter if you focus on solving for maximum value of $x + y$ so don't actually need to calculate $a,b,c,d$ individually.
Additional hint:
Can you see why to get the maximum value of $x + y$ then $x + y = b + c + 2d$?
answered Jan 25 at 14:37
DubsDubs
55926
answered Jan 25 at 14:37
DubsDubs
55926
answered Jan 25 at 14:37
DubsDubs
55926
answered Jan 25 at 14:37
DubsDubs
55926
55926
add a comment |
add a comment |
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$begingroup$
Sounds good to me
$endgroup$
– lioness99a
Jan 25 at 14:11
$begingroup$
@lioness99a so it's correct right?
$endgroup$
– S.Nep
Jan 25 at 14:12
$begingroup$
Without working through the equations myself I can't be 100% sure but the method sounds alright
$endgroup$
– lioness99a
Jan 25 at 14:19
$begingroup$
Are you sure it is 793, I get 761 as the maximum.
$endgroup$
– Satish Ramanathan
Jan 25 at 14:41
1
$begingroup$
Agree with Satish, I'm getting 761 as well.
$endgroup$
– Dubs
Jan 25 at 14:46