| Vicky Neale |
The following is the discussion about the 2003 BMO2 Question 1, which was: For each integer n > 1, let p(n) denote the largest prime factor of n . Determine all triples x ,y ,z of distinct positive integers satisfying (i) x ,y ,z are in arithmetic progression, and (ii) p (xyz ) < = 3. |
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| Chris
Tynan |
For q.1 I get this: Since p(xyz) < 4 then p(xyz) = 2 or 3 Therefore, xyz = 2a or 3b or 2a .3b However, it is impossible by xyz to be 2a or 3b since this would mean x,y,z are all powers of 2 or 3 respectively, but they can't be, since that would be GP. Therefore xyz = 2a .3b Also, x + y + z = 3y (AP) Saying x = 2n .3m call y = 2x and z = 3x The product will only be divisible by 2 and 3 since the product is 6 x3 and x is only divisible by 2 and 3 - This gives an infinite family of solutions. Question 2 looks OK, but 4 looks quite hard... Chris |
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| Peter
Conlon |
For q.1: Chris, you've got an infinite family of solutions, but not all of them. For example, x=2n .3m , y=2n-1 .3m+1 , z=2n+1 .3m is a different family, but still has the desired properties. Peter |
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| Shahnawaz
Abdullah |
well, here's my soln for 1: x= x and of form (2^a)(3^b) y = x+d and of form (2^c)(3^d) z = x + 2d and of form (2^e)(3^f) 2d = (2^e)(3^f) - (2^a)(3^b) which implies d is multiple of 2 and of 3. As a result, we only need to consider d = 1 for the base triplet of the families of triplets. Therefore: x=x y=x+1 z=x+2 since all these are mults of 2 and 3 with no other prime factors there is a max of 1 multiple of 3 term there and a max of 2 non consecutive terms which are mults of 2 Therefore, only base family is 0,1,2 1,2,3 2,3,4 and all other triplets can be formed by multiplying EVERY term by the same factor of (2^n)(3^m) where n and m are positivve integers |
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| Kerwin
Hui |
I'm a bit surprised to learn on Monday evening (from one of the BMO2 setter) that there are no inequalities this year. This breaks the usual rule of 1 inequality, 1 geometry, 1 combinatorical (usually dwarfs) and 1 on something else... Shahnawaz, I'm afraid your solution cannot worth anything more than 3/10 as it stands. There are large gaps that need filling (e.g. 2d=2e 3f -2a 3b implies d is a multiple of 2 and 3 is just false, and you haven't really justify why it suffices to consider only d=1 - you have to prove that d divides x). Kerwin |
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| David
Turton |
I'm a bit stuck at the end of Q1... Like Shahnawaz said, if we have a base triple then all multiples of each term by 2n 3m are also triples which work. So we only need consider cases where gcd(x,y,z)=1. I've managed to show that when gcd(x,y,z)=1, x must be 1 or 2 and I'm down to two cases: Case 1: (1,2a ,3b ) ...being equivalent to solving 2a+1 =3b +1. I suspect a=b=1 giving (1,2,3) is the only solution but can't seem to get anywhere. Case 2: (2,3c ,2d ) ...being equivalent to solving 2d-1 =3c -1. If c is even, c can only be 2 giving (2,9,16), but I'm a bit stumped for c odd. I suspect if c odd, c can only be 1 giving (2,3,4) but I can't be sure there aren't more solutions. Any ideas? Otherwise, Q2 was nice (Peter I did it exactly your way), Q3 I don't seem to be able to get started on though. A hint would be nice... Thanks Dave. |
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| Chris
Tynan |
David, why does x neccessarily have to be 1 or 2? Also, how come gcd(x,y,z) = 1? Surely x,y,z will have a common factor of 2 and/or 3? Take for instance, the counterexample (12,18,24), where gcd(x,y,z) = 6 Chris |
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| David
Turton |
Chris, Maybe I wasn't clear enough... our families of triples will all consist of a base triple (where gcd(x,y,z)=1) with each term multiplied by a factor of 2n 3m . The example you presented is one of the family with base triple (2,3,4); the example in your first post is of the form with base (1,2,3). To determine all families of solutions we need to find all base triples, i.e. considering cases where gcd(x,y,z)=1. Given this, try using the fact that x,y,z are in arithmetic progression to narrow down possibilities for the x-value of a base triple. Dave |
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| Chris
Tynan |
OK, I see now - I understand your answer. Thanks Chris |
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| David
Turton |
Well I wouldn't call it an answer just yet... I've proved that in case 1, a=b=1 is in fact the only solution, and that in case 2, when c is odd, d is even. So solving 2d-1 =3c -1 for d even and c odd will finish it! So close... Dave |
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| Vicky Neale |
At this point Michael Doré advised working mod 4 and David was able to finish the question. |