wow, this is way more challenging than A1, and I didn't find A1 easy, either. Despite your great presentation and clear explanations, it's still hard to grasp. No wonder this test is considered so difficult
When you have the term 75+b/2 in the second coordinate (in the b not equal to 50 case) what is stopping b from being 52 or more making the second coordinate invalid?
Short answer is because the Putnam 2018-B1 was explicitely asking for partitions into two sets. Long answer is because P-{v} has 302 elements, and 302 = 2 x 151 (where 151 is prime). Therefore P-{v} can have partitions with equal size only into 2 sets or into 151 sets. Now, if you follow the same explanation as in the video up to 3:40 (just replacing 2SigmaA by 151sigmaA) and then compare the sum of P with 151SigmaA + (a,b) mod 151 (instead of mod 2), you should have "a" congruent to 1 mod 151 (and therefore = 1 because it belongs to {0,1,2}), and "b" congruent to 50 mod 151 (and therefore = 50 because it belongs to {0,1,2,3,...,100}). So, in this case the only possible v is (1,50) and you can split P-{v} into 151 pairs : {(0,k),(2,100-k)} (with k varying from 0 to 100 : 101 pairs) and {(1,k),(1,100-k)} (with k varying from 0 to 49 : 50 pairs). As you can see, each pair adds up to (2,100). Indeed there is also a solution splitting P-{v} into 151 pairs instead of 2 sets of 151 elements each. But in this case the only possible v is (1,50).
Neat problem. I solved it in a slightly different way. When I came back to rewatch I panicked for a sec, when I saw that the thumbnail has a different problem - a should go up to 2 and not 1 (that makes for a more boring problem :) )
I think my method was pretty easy. I worked out that it had to be (1, even) based on the sums mod 2. Partitioning the 0s, 1s and 2s is easy. Now focus on the 0 to 100 part and take an even number out. You have 3 lots of 0 to 100 with just one number missing. To equally divide it, put 0 to 100 on one side and 0 to 100 on the other side. Then you have 100 numbers left (0 to 100 with one even number missing). You try to divide those pretty evenly, like the top goes in group A, the next goes to group B, then back to group A, etc. All that matters is that the sum ends up pretty close, less than 200 apart (they'll be about 50 apart). Let's say the difference is 68. Take 34 from the larger group and swap it with 0 from the smaller group. Done.
6:34 no idea whats going on, this is far above any math i know but it should be (1 51) not (1 59)
Unless im wrong
You are right, it needs to be (1 51) so that the sum of that pair is (2 100).
wow, this one went straight over my head.
wow, this is way more challenging than A1, and I didn't find A1 easy, either. Despite your great presentation and clear explanations, it's still hard to grasp. No wonder this test is considered so difficult
Who else thought the (a [new line] b) in the top left meant a choose b and then got confused why a’s max value was 2 and b’s was 100?
Really nice. If we take 0
When you have the term 75+b/2 in the second coordinate (in the b not equal to 50 case) what is stopping b from being 52 or more making the second coordinate invalid?
He wrote 75-b/2, not 75+b/2. It will fall between 25 and 75, which are valid values.
Wait why can't there be a partition into 3, 4, 5, etc sets with equal size and sum?
Short answer is because the Putnam 2018-B1 was explicitely asking for partitions into two sets.
Long answer is because P-{v} has 302 elements, and 302 = 2 x 151 (where 151 is prime). Therefore P-{v} can have partitions with equal size only into 2 sets or into 151 sets. Now, if you follow the same explanation as in the video up to 3:40 (just replacing 2SigmaA by 151sigmaA) and then compare the sum of P with 151SigmaA + (a,b) mod 151 (instead of mod 2), you should have "a" congruent to 1 mod 151 (and therefore = 1 because it belongs to {0,1,2}), and "b" congruent to 50 mod 151 (and therefore = 50 because it belongs to {0,1,2,3,...,100}).
So, in this case the only possible v is (1,50) and you can split P-{v} into 151 pairs : {(0,k),(2,100-k)} (with k varying from 0 to 100 : 101 pairs) and {(1,k),(1,100-k)} (with k varying from 0 to 49 : 50 pairs). As you can see, each pair adds up to (2,100). Indeed there is also a solution splitting P-{v} into 151 pairs instead of 2 sets of 151 elements each. But in this case the only possible v is (1,50).
Neat problem. I solved it in a slightly different way. When I came back to rewatch I panicked for a sec, when I saw that the thumbnail has a different problem - a should go up to 2 and not 1 (that makes for a more boring problem :) )
I think my method was pretty easy. I worked out that it had to be (1, even) based on the sums mod 2. Partitioning the 0s, 1s and 2s is easy.
Now focus on the 0 to 100 part and take an even number out. You have 3 lots of 0 to 100 with just one number missing. To equally divide it, put 0 to 100 on one side and 0 to 100 on the other side. Then you have 100 numbers left (0 to 100 with one even number missing). You try to divide those pretty evenly, like the top goes in group A, the next goes to group B, then back to group A, etc. All that matters is that the sum ends up pretty close, less than 200 apart (they'll be about 50 apart). Let's say the difference is 68. Take 34 from the larger group and swap it with 0 from the smaller group. Done.
But what is it good for?
you mean math?
idk pretty useful
What a boring problem!
what a boring comment!
Glenn Wouda LoL
okay karen