Prove that it will always be possible to find such two weights so the weight difference between them wouldn't...
$begingroup$
You have 40 weights. It is known that the difference of weight in every 2 weights is no bigger than 45 kg. Also, it is known that you can divide every single group of 10 weights into 2 groups (5 weights in each) and the sum of weights in these 2 groups won't differ by more than 11 kg. You have to prove that it will always be possible to find such two weights so the weight difference between them wouldn't be bigger than 1 kg.
What I got is that our all weights must be between (and including) $n$ kg and $n+45$ kg. But I'm not sure should I use the fact, that you can divide every single group of 10 weights into 2 groups (5 weights in each) and the sum of weights in these 2 groups won't differ by more than 11 kg.
Any hint would be very appreciated.
proof-writing
asked Dec 16 '18 at 16:34
thomas21thomas21
159112
$endgroup$
add a comment |
$begingroup$
You have 40 weights. It is known that the difference of weight in every 2 weights is no bigger than 45 kg. Also, it is known that you can divide every single group of 10 weights into 2 groups (5 weights in each) and the sum of weights in these 2 groups won't differ by more than 11 kg. You have to prove that it will always be possible to find such two weights so the weight difference between them wouldn't be bigger than 1 kg.
What I got is that our all weights must be between (and including) $n$ kg and $n+45$ kg. But I'm not sure should I use the fact, that you can divide every single group of 10 weights into 2 groups (5 weights in each) and the sum of weights in these 2 groups won't differ by more than 11 kg.
Any hint would be very appreciated.
proof-writing
asked Dec 16 '18 at 16:34
thomas21thomas21
159112
$endgroup$
$begingroup$
Do the weights take integral values only?
$endgroup$
– Sameer Baheti
Dec 16 '18 at 17:05
$begingroup$
Not necessarily.
$endgroup$
– thomas21
Dec 16 '18 at 18:03
add a comment |
$begingroup$
You have 40 weights. It is known that the difference of weight in every 2 weights is no bigger than 45 kg. Also, it is known that you can divide every single group of 10 weights into 2 groups (5 weights in each) and the sum of weights in these 2 groups won't differ by more than 11 kg. You have to prove that it will always be possible to find such two weights so the weight difference between them wouldn't be bigger than 1 kg.
What I got is that our all weights must be between (and including) $n$ kg and $n+45$ kg. But I'm not sure should I use the fact, that you can divide every single group of 10 weights into 2 groups (5 weights in each) and the sum of weights in these 2 groups won't differ by more than 11 kg.
Any hint would be very appreciated.
proof-writing
asked Dec 16 '18 at 16:34
thomas21thomas21
159112
$endgroup$
You have 40 weights. It is known that the difference of weight in every 2 weights is no bigger than 45 kg. Also, it is known that you can divide every single group of 10 weights into 2 groups (5 weights in each) and the sum of weights in these 2 groups won't differ by more than 11 kg. You have to prove that it will always be possible to find such two weights so the weight difference between them wouldn't be bigger than 1 kg.
What I got is that our all weights must be between (and including) $n$ kg and $n+45$ kg. But I'm not sure should I use the fact, that you can divide every single group of 10 weights into 2 groups (5 weights in each) and the sum of weights in these 2 groups won't differ by more than 11 kg.
Any hint would be very appreciated.
proof-writing
proof-writing
asked Dec 16 '18 at 16:34
thomas21thomas21
159112
asked Dec 16 '18 at 16:34
thomas21thomas21
159112
asked Dec 16 '18 at 16:34
thomas21thomas21
159112
asked Dec 16 '18 at 16:34
thomas21thomas21
159112
asked Dec 16 '18 at 16:34
thomas21thomas21
159112
159112
$begingroup$
Do the weights take integral values only?
$endgroup$
– Sameer Baheti
Dec 16 '18 at 17:05
$begingroup$
Not necessarily.
$endgroup$
– thomas21
Dec 16 '18 at 18:03
add a comment |
$begingroup$
Do the weights take integral values only?
$endgroup$
– Sameer Baheti
Dec 16 '18 at 17:05
$begingroup$
Not necessarily.
$endgroup$
– thomas21
Dec 16 '18 at 18:03
$begingroup$
Do the weights take integral values only?
$endgroup$
– Sameer Baheti
Dec 16 '18 at 17:05
$begingroup$
Do the weights take integral values only?
$endgroup$
– Sameer Baheti
Dec 16 '18 at 17:05
$begingroup$
Not necessarily.
$endgroup$
– thomas21
Dec 16 '18 at 18:03
$begingroup$
Not necessarily.
$endgroup$
– thomas21
Dec 16 '18 at 18:03
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Just an idea...see if it can be made to work.
Proof by contradiction: for any two weights $w_i, w_j, |w_i-w_j|>1$. Put them all in ascending order. Without loss of generality, we can probably start with $epsilon >0$ as the smallest weight. Then take the lightest 5 weights and the heaviest 5 weights - can they still be partitioned so that the weight difference is less than 11? That would be a contradiction to "every single group of 10 weights into 2 groups (5 weights in each) and the sum of weights in these 2 groups won't differ by more than 11 kg."
answered Dec 17 '18 at 1:32
MatthiasMatthias
3287
$endgroup$
$begingroup$
solved it, thank you very much
$endgroup$
– thomas21
Dec 18 '18 at 16:15
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Just an idea...see if it can be made to work.
Proof by contradiction: for any two weights $w_i, w_j, |w_i-w_j|>1$. Put them all in ascending order. Without loss of generality, we can probably start with $epsilon >0$ as the smallest weight. Then take the lightest 5 weights and the heaviest 5 weights - can they still be partitioned so that the weight difference is less than 11? That would be a contradiction to "every single group of 10 weights into 2 groups (5 weights in each) and the sum of weights in these 2 groups won't differ by more than 11 kg."
answered Dec 17 '18 at 1:32
MatthiasMatthias
3287
$endgroup$
$begingroup$
solved it, thank you very much
$endgroup$
– thomas21
Dec 18 '18 at 16:15
add a comment |
$begingroup$
Just an idea...see if it can be made to work.
Proof by contradiction: for any two weights $w_i, w_j, |w_i-w_j|>1$. Put them all in ascending order. Without loss of generality, we can probably start with $epsilon >0$ as the smallest weight. Then take the lightest 5 weights and the heaviest 5 weights - can they still be partitioned so that the weight difference is less than 11? That would be a contradiction to "every single group of 10 weights into 2 groups (5 weights in each) and the sum of weights in these 2 groups won't differ by more than 11 kg."
answered Dec 17 '18 at 1:32
MatthiasMatthias
3287
$endgroup$
$begingroup$
solved it, thank you very much
$endgroup$
– thomas21
Dec 18 '18 at 16:15
add a comment |
$begingroup$
Just an idea...see if it can be made to work.
Proof by contradiction: for any two weights $w_i, w_j, |w_i-w_j|>1$. Put them all in ascending order. Without loss of generality, we can probably start with $epsilon >0$ as the smallest weight. Then take the lightest 5 weights and the heaviest 5 weights - can they still be partitioned so that the weight difference is less than 11? That would be a contradiction to "every single group of 10 weights into 2 groups (5 weights in each) and the sum of weights in these 2 groups won't differ by more than 11 kg."
answered Dec 17 '18 at 1:32
MatthiasMatthias
3287
$endgroup$
Just an idea...see if it can be made to work.
Proof by contradiction: for any two weights $w_i, w_j, |w_i-w_j|>1$. Put them all in ascending order. Without loss of generality, we can probably start with $epsilon >0$ as the smallest weight. Then take the lightest 5 weights and the heaviest 5 weights - can they still be partitioned so that the weight difference is less than 11? That would be a contradiction to "every single group of 10 weights into 2 groups (5 weights in each) and the sum of weights in these 2 groups won't differ by more than 11 kg."
answered Dec 17 '18 at 1:32
MatthiasMatthias
3287
answered Dec 17 '18 at 1:32
MatthiasMatthias
3287
answered Dec 17 '18 at 1:32
MatthiasMatthias
3287
answered Dec 17 '18 at 1:32
MatthiasMatthias
3287
3287
$begingroup$
solved it, thank you very much
$endgroup$
– thomas21
Dec 18 '18 at 16:15
add a comment |
$begingroup$
solved it, thank you very much
$endgroup$
– thomas21
Dec 18 '18 at 16:15
$begingroup$
solved it, thank you very much
$endgroup$
– thomas21
Dec 18 '18 at 16:15
$begingroup$
solved it, thank you very much
$endgroup$
– thomas21
Dec 18 '18 at 16:15
add a comment |
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$begingroup$
Do the weights take integral values only?
$endgroup$
– Sameer Baheti
Dec 16 '18 at 17:05
$begingroup$
Not necessarily.
$endgroup$
– thomas21
Dec 16 '18 at 18:03