27 May 09
I'll be posting a few math contest problems a friend sent me (I believe these are early highschool problems, but some are quite challenging). Here's no.1!
1. Winnie Pooh and Piglet left their houses at the same time to
visit one another. Due to the fog they did not notice one another. Piglet
reached Winnie Pooh’s place in four minutes and Winnie Pooh reached
Piglet’s place one minute after they “met”. How long was everybody on
his way?
27 May 09
Originally posted by PBE6Unless I am missing something, I get that Piglet was on his way for 4 minutes (as given) and Pooh was on his way for [1+sqrt(17)]/2 minutes, or ~2.56 minutes. This is under the further assumptions that they both walked the same distance (on parallel paths) and that each walked at constant speed.
I'll be posting a few math contest problems a friend sent me (I believe these are early highschool problems, but some are quite challenging). Here's no.1!
1. Winnie Pooh and Piglet left their houses at the same time to
visit one another. Due to the fog they did not notice one another. Piglet
reached Winnie Pooh’s place in four minutes and Winnie Pooh reached
Piglet’s place one minute after they “met”. How long was everybody on
his way?
28 May 09
Originally posted by PBE6i don get it..is it "Piglet
I'll be posting a few math contest problems a friend sent me (I believe these are early highschool problems, but some are quite challenging). Here's no.1!
1. Winnie Pooh and Piglet left their houses at the same time to
visit one another. Due to the fog they did not notice one another. Piglet
reached Winnie Pooh’s place in four minutes and Winnie Pooh reached
Piglet’s place one minute after they “met”. How long was everybody on
his way?
reached Winnie Pooh’s place in four minutes AFTER THEY "MET" and Winnie Pooh reached
Piglet’s place one minute AFTER THEY "MET" ?? i mean the time is given from there meeting place in both instances right? this makes the above answer wrong..:-(
28 May 09
1 edit
Originally posted by blacknight1985Oops, you may be right, and I may have misinterpreted the problem statement.
i don get it..is it "Piglet
reached Winnie Pooh’s place in four minutes AFTER THEY "MET" and Winnie Pooh reached
Piglet’s place one minute AFTER THEY "MET" ?? i mean the time is given from there meeting place in both instances right? this makes the above answer wrong..:-(
I think in this case the problem is easier, and a solution would be that Piglet was on his way for 6 minutes; and Pooh was on his way for 3 minutes. They "met" after 2 minutes when Pooh was 2/3 the way on his respective journey and Piglet was 1/3 the way on his respective journey.
28 May 09
Originally posted by LemonJelloCalculations people, answers don't give much/any credit 🙂
Oops, you may be right, and I may have misinterpreted the problem statement.
I think in this case the problem is easier, and a solution would be that Piglet was on his way for 6 minutes; and Pooh was on his way for 3 minutes. They "met" after 2 minutes when Pooh was 2/3 the way on his respective journey and Piglet was 1/3 the way on his respective journey.
28 May 09
Originally posted by PBE6ok here it is...take t = 0 when they left their respective houses..assume at t = t1 they 'meet'.. take the total distance between the 2 places as S and the 'meeting' place is at x distance from Pooh's house..so we have,
Very true! Let's see those calculations. 🙂
Velocity(Pooh) = x/t1 = (S-x)/1 --->(1)
Velocity(Pig) = (S-x)/t1 = x/4 --->(2)
dividing (1) by (2) so that (S-x) and x terms cancel out, we get t1 = 2 mins..
sooo Pooh takes t1+1 = 3 mins
and Pig takes t1+4 = 6 mins.. ta da 😀
u can also prove x = 2S/3
cheers!
28 May 09
Originally posted by blacknight1985Excellent!
ok here it is...take t = 0 when they left their respective houses..assume at t = t1 they 'meet'.. take the total distance between the 2 places as S and the 'meeting' place is at x distance from Pooh's house..so we have,
Velocity(Pooh) = x/t1 = (S-x)/1 --->(1)
Velocity(Pig) = (S-x)/t1 = x/4 --->(2)
dividing (1) by (2) so that (S-x) and x terms can ...[text shortened]... t1+1 = 3 mins
and Pig takes t1+4 = 6 mins.. ta da 😀
u can also prove x = 2S/3
cheers!
Here's another one:
Problem 2 (MathBattle) It is known that m^2 + n^2 + m is divisible by mn
for some positive integers m and n. Prove that m is a perfect square.
28 May 09
Originally posted by blacknight1985um, the question says, "Piglet reached Winnie Pooh’s place in four minutes"
ok here it is...take t = 0 when they left their respective houses..assume at t = t1 they 'meet'.. take the total distance between the 2 places as S and the 'meeting' place is at x distance from Pooh's house..so we have,
Velocity(Pooh) = x/t1 = (S-x)/1 --->(1)
Velocity(Pig) = (S-x)/t1 = x/4 --->(2)
dividing (1) by (2) so that (S-x) and x terms can ...[text shortened]... t1+1 = 3 mins
and Pig takes t1+4 = 6 mins.. ta da 😀
u can also prove x = 2S/3
cheers!
So how could it take him 6 minutes?
29 May 09
Originally posted by PBE6Yikes, this one was a lot tougher than the Piglet/Pooh question!! It has taken me a while, but I believe that I have a proof (by contradiction). It might be hard for me to summarize it here clearly, but whatever, here goes:
Excellent!
Here's another one:
Problem 2 (MathBattle) It is known that m^2 + n^2 + m is divisible by mn
for some positive integers m and n. Prove that m is a perfect square.
Suppose it is not the case that m is a perfect square. Then, there must be at least one odd power in the prime-power factorization of m. This means there must exist some prime number P such that P^(2a+1) exactly divides m where a is some element of the set [0,1,2,3,....]. (Since positive integer m is a perfect square if and only if all powers in the prime-power factorization of m are even.***) By "exactly" divides I mean that P^(2a+1) divides m but P^(2a+2) does not.
Since we know that m^2 + n^2 + m = bmn (for some positive integer b), it is clear that m divides n^2. Since m divides n^2; and since P^(2a+1) divides m; it follows that P^(2a+1) divides n^2. But, it further is the case that P^(2a+2) divides n^2 because n^2 is a perfect square and because, again, such a perfect square must have all even powers in its prime-power factorization*** (here, I am using the idea that if P^k divides a perfect square integer where k is an odd positive integer then it must be that P^(k+1) also divides the same perfect square). It follows that P^(a+1) divides n. Of course, P^(a+1) also divides m.
Since P^(a+1) divides both m and n, it follows that P^(2a+2) divides mn. Since bmn = m^2 + n^2 + m, it follows that P^(2a+2) divides m^2 + n^2 + m. Since P^(2a+2) divides both m^2 and n^2, it follows that P^(2a+2) also divides m. But from above, P^(2a+1) exactly divides m, which entails that P^(2a+2) does not divide m. So we have a contradiction.
So m is a perfect square.
-------------------------------
***Since a couple parts of the proof hinge on this, I will show the basic idea for proof of this here:
Suppose that n has all even powers in its prime-power factorization. Then we could write n = P1^(2c1)*P2^(2c2)*...*Pk^(2ck). In that case, n = m^2 where m = P1^(c1)*P2^(c2)*...*Pk^(ck). So n is a perfect square. To show the other way, suppose n is a perfect square, such that n = m^2. The integer m will have some prime-power factorization m = P1^(c1)*P2^(c2)*...*Pk^(ck). Then n has prime-power factorization n = P1^(2c1)*P2^(2c2)*...*Pk^(2ck). So n has all even powers in its factorization. So positive integer n is a perfect square if and only if all powers in the prime-power factorization of n are even.
04 Jun 09
Originally posted by LemonJellothis is very nice... some of the language could potentially be cleared up, but all in all your proof is very solid! i wonder if the writer of the question knew it would be that hard, or if they stopped at m|n^2 therefore m is a perfect square (clearly untrue).
Yikes, this one was a lot tougher than the Piglet/Pooh question!! It has taken me a while, but I believe that I have a proof (by contradiction). It might be hard for me to summarize it here clearly, but whatever, here goes:
Suppose it is not the case that m is a perfect square. Then, there must be at least one odd power in the prime-power factorizat ...[text shortened]... is a perfect square if and only if all powers in the prime-power factorization of n are even.
or perhaps there's a really nice quick elegant solution stemming m|n^2 and n|(m^2 + m)... i've been looking for a parity argument or a modular arithmetic solution, but there are too many possibilities and seemingly no outright contradictions (other than your method) from assuming m to not be a perfect square.
anyone else have cool solutions? 🙂
04 Jun 09
Originally posted by AetheraelThanks. I tried other approaches, but I found them to be dead ends. I don't see a proof other than the one I gave, but maybe someone else can come up with a more elegant proof.
this is very nice... some of the language could potentially be cleared up, but all in all your proof is very solid! i wonder if the writer of the question knew it would be that hard, or if they stopped at m|n^2 therefore m is a perfect square (clearly untrue).
or perhaps there's a really nice quick elegant solution stemming m|n^2 and n|(m^2 + m)... i'v ...[text shortened]... our method) from assuming m to not be a perfect square.
anyone else have cool solutions? 🙂
04 Jun 09
Originally posted by LemonJelloVery impressive 😀
Yikes, this one was a lot tougher than the Piglet/Pooh question!! It has taken me a while, but I believe that I have a proof (by contradiction). It might be hard for me to summarize it here clearly, but whatever, here goes:
Suppose it is not the case that m is a perfect square. Then, there must be at least one odd power in the prime-power factorizat ...[text shortened]... is a perfect square if and only if all powers in the prime-power factorization of n are even.
I wonder though, is the contradiction nessecary? Your proof looks like it would work without the contradiction. Though I don't ahve the time to really look at it at this moment 🙂
04 Jun 09
1 edit
Originally posted by TheMaster37i definitely think it's necessary... his whole proof hinges on an assumption that in the prime factorization of m there is some prime p that has an odd multiplicity. but then since m|n^2 he can show an even multiplicity of p in n. then later (after some very nice work, i might add) he shows that p in fact has an even multiplicity in m and therefore his initial assumption must have been false.
Very impressive 😀
I wonder though, is the contradiction nessecary? Your proof looks like it would work without the contradiction. Though I don't ahve the time to really look at it at this moment 🙂
it doesn't look like a direct proof is possible with this line of reasoning because of the step where since p^2k+1 divides n that p^2k+2 also divides n, which entirely stems from his assumption, and is the lynchpin that leads to the contradiction.
i could be wrong though, and would LOVE to see a direct proof! 🙂