Mathematical Problem Solving TEAM A-11/12/2010

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1 Mathematical Problem Solving TEAM A-/2/200 Tramaine Find all integer solutions to x 2 + y 2 + z 2 = 2xyz. Solution. We see there is no solution other than x = y = z = 0. In fact, assume x, y, z are integers, not all 0, solving the equation. Then the left hand side is positive, thus the right hand side can t be 0; none of the integers can be 0. Assume they are all even; say x = 2m, y = 2n, z = 2k, where m, n, k are integers. Then 4m 2 4n 2 +4k 2 = 8mnk, hence m 2 +n 2 +k 2 = 2mnk, and m, n, k are also solutions of the equation. If we keep dividing by 2, we will eventually obtain a solution in which one of the elements is odd. This proves that if there is a solution with not all elements equal to 0, there is one with one of the elements an odd integer; tat is, there exists a solution x, y, z such that one of x, y, z is odd. Without loss of generality we may assume that x is odd. Since the right hand side of the equation is even, we conclude that there has that one of y, z is odd, the other one even; without loss of generality we can assume y is odd and z is even. Since z is even, 2xyz 0 (mod 4). However, mod 4 the squares of all odd integers are, of even integers are 0, thus x 2 + y 2 + z = 2 (mod 4). Thus contradicting the existence of a non-zero solution. 2 Hai x 2 + y 2 + z xyz (mod 4), For a deck containing an even number of cards define a perfect shuffle as follows: Divide the deck in two equal halves, the top half and the bottom half; then interleave the cards one by one between the two halves, starting with the top card of the bottom half, then the top card of the bottom half, etc. Determine the minimum number of perfect shuffles needed to restore a 94-card deck to its original order. Can you generalize this to decks of arbitrary (even) size? Solution. Suppose the deck contains n = 2m cards. We define a function f : {, 2,..., 2m} {, 2,..., 2m} as follows: If a card is in position k, then f(k) is the position in which it lands after a perfect shuffle. So if a card is originally in position k, after one shuffle it is in position f(k), after two shuffles in position f(f(k)), and so forth. To determine an expression for f we work mod n + = 2m +. (YES: 2m +, NOT 2m ). We see that every card from the bottom half; the cards numbered,..., m, goes to a position twice its original position: In the first shuffle (for example) goes to second place, 2 to fourth place,..., m to the last place; that is to position 2m. The card in position m +, the top of the bottom half, goes to position, m + 2 to 3, etc. Another way of saying this is to say that the card that was in position m + ends in position 2(m + ) (2m + ) =, the one in position 2m + 2 moves to position 2(2m + 2) (2m + ) = 3; etc. Briefly: { 2k if k m, f(k) = 2k (2m + ) if m + k 2m. Even more briefly: f(k) 2k(mod 2m + ). Thus shuffling moves every card to a new position that is twice the original one, reduced mod 2m +. After l shuffles, the card that was originally in position k is in position 2 l k. The minimum number of shuffles it takes to get everything back to where it was at first is thus the smallest positive integer l such that 2 l mod 2m +. This number is called the order of 2 mod 2m +. To determine it (up to a point), let us recall Euler s Theorem. Since gcd(2, 2m + ) =, we see that 2 ϕ(2m+) (mod 2m + ). This shows that l ϕ(2m + ). We can say a bit more. We can divide ϕ(2m + ) by l to get ϕ(2m + ) = al + r for some integers a and r, 0 r < l. Then 2 ϕ(2m+) = ( 2 l) a 2 r a 2 r = 2 r

2 hence 2 r (mod 2m + ). But l was the smallest positive integer such that 2 raised to that power was congruent to mod 2m + and 0 r < l, thus r = 0: l divides ϕ(2m + ). This is the best we can say in the general case: The smallest number of shuffles needed to restore a 2m-deck to its original order is a divisor of ϕ(2m + ). It has to be a divisor >, of course. For the case of n = 94 we need to find the order of 2 mod 95. Now 95 = 5 9, thus ϕ(95) = 95( /5)( /9) = 72. The divisors of 72 are 2, 4, 6, 8, 9, 2, 8, 24, 36, 72. The first of these powers of 2 larger than 95 is 8; Now (with meaning always mod 95) 2 8 = (mod 95) , 2 8 = ( 2 9) = , 2 36 = ( 2 8) = 52. The order is 36. Any potential smaller order would have to divide 36. Of the divisors of 36 the only one we haven t checked yet is 2; it is easy to see it won t work. 3 Ivan (a) Show that if 2 k is a prime, then 2 k (2 k ) is perfect. (b) Prove that if n is even and perfect, then n = 2 k (2 k ), where 2 k is prime. Solution. We denote by σ(n) the sum of all the divisors of n. Thus n is perfect if and only if σ(n) = 2n. (a) Let p = 2 k and assume it is prime. We have to show that n = 2 k p is perfect. Because p is prime, the divisors of n are (precisely!), 2,..., 2 k, p, 2p,..., 2 k p so that σ(n) = ( k ) + ( + 2,..., 2 k )p = ( + 2,..., 2 k )(p + ). If we use the formula for sum of a geometric progression + 2,..., 2 k = 2 k and the fact that p = 2 k, we see that This proves n perfect. σ(n) = (2 k )(2 k + ) = p2 k = 2(2 k p) = 2n. (b) Assume that n is even and perfect. To prove that it has the desired form, we have to use some properties of the function σ, specially that it is multiplicative: σ(nm) = σ(n)σ(m) if gcd(n, m) =. It is easy to calculate σ(n) if n is the power of a prime; if n = p k the divisors of n are precisely the integers, p,..., p k and by the formula for sum of a geometric progression, σ(p k ) = pk+ p. Assume now n is even and perfect. Because it is even, the prime power decomposition of n looks like n = 2 k p e per r, where k (because n is even) and p,..., p r are distinct odd primes; e,..., e r.by perfection and multiplicativity of σ, pe r r = 2n = σ(n) = σ(2 k )σ(p e ) σ(pe r ) = (2 k+ ) j= p. If we look at these equations carefully, specifically the equality of the left hand side of the first equation with the right hand side of the last one; that is at per r = (2 k+ ) p, () j= 2

3 we might notice how tight it is. (Incidentally, comments like these are NOT part of a proof and can appear in writings of an instructor or in a textbook, but should probably not appear in writings by students. Except if the student is acting as an instructor, which can happen at times.) Notice that p ej+ = p ej j + p ej j + + > p ej j (2) p so all the factors on the righthand side are strictly greater than a corresponding factor on the left hand side, except for the first factor, which is 2 k+ on the right hand side and 2 k+ on the left hand side. But 2 k+ is not so much smaller than 2 k+, specially if k is large. Is this difference enough to make up for how much larger all the other factors can be? The answer turns out to be: barely. Let us return to the proof as such. The key idea is to realize that 2 k+ must be divisible by one of the p j s because it is odd. In equation (), the left hand side must be the prime power decomposition of the right hand side; every prime factor of the right hand side appears on the left hand side. Since 2 k+ is odd, it must be divisible by one of the p j s; without loss of generality we may assume it is p. Then p 2 k+, hence /p /(2 k+ ), +/p +/(2 k+ ) = 2 k+ /(2 k+ ) and ) ) ( + p p e+ ( = p e p + pe + + = p e + p + + p e p e Using this information in () gives us the following interesting inequality (or inequalities): per r = (2 k+ ) j= p (2 k+ )p e 2 k+ 2 k+ j=2 p e = p 2 k+ 2 k+. (3) j=2 p. This is almost a contradiction. If r 2; if there are as many as two odd primes in the prime power decomposition of n, we have a contradiction. In fact, using (2) for j = 2,..., r gives and using this in the interesting inequality gives j=2 > p e 2 2 p pe r r per r > per r, which is nonsense. We conclude that r = ; n = 2 k p e for a single odd prime p. Let us return to equation (). which now reads ( = (2k+ ) pe + = (2 k+ )p e ). (4) p p p e Recall that we had p 2 k+ and from this we concluded that + p + + p e If ANY of these inequalities is strict, the end result would be + p 2k+ 2 k+. 2 k+ p e > (2k+ )p e 2k+ 2 k+ = 2 k+ p e = 2 k+ > 2 k+, which is again nonsense. Thus all these inequalities must be equalities, in particular, + p + + p e = + p, only possible if e =, and p = 2 k+. We proved that n = 2 k (2 k+ ) where 2 k+ is prime. We are done. 3

4 4 Mayra Let m and n be arbitrary non-negative integers. Show that (2m)!(2n)! m!n!(m + n)! is an integer. Solution. Looking up the online hints page, one finds as a hint to count powers of primes in numerator and denominator. Here is a proof doing just that. I will divide into several parts so it is easier to read (or, I hope it will be easier to read, assuming anybody reads it). Step. Let a, b be positive integers, let p be prime and for r =, 2,... let k r be the number of multiples of p r that are larger than or equal a and less than or equal a + b. Then the highest power of p dividing the product a(a + ) (a + b) is r= k r. Before proving this, let me try to clear it up with an example. Say a = 7, b = 25, p = 3. The product we are looking at is The multiples of 3 in that product are 9, 2, 5, 8, 2, 24, a total of six; thus k = 6. The multiples of 3 2 = 9 are 9, 8, thus k 2 = 2. There are no multiples of 3 3 = 27, 3 4 = 8, etc.; thus k r = 0 if r 3 and the highest power of 3 dividing the product in question is = 8. It is easy to see that the product in question equals showing the correctness of the result in this case. In general, if r is large enough, then p r > a + b, and k r = 0. The sum r= k r is always a finite sum. The proof is fairly obvious. Every multiple of p provides one power of p. Multiples of p 2 provide an additional power. If there is a multiple of p 3 in the list, each such multiple provides an additional power. And so forth. That s about it. Step 2. Let a, b be integers, a 0, b, let p be prime, let r N, and let k r be the number of multiples of p r that are larger than or equal a + and less than or equal a + b. Then k r = a + b p r a p r. Proof. Let j be the first integer such that jp r is larger than or equal to a; that is, ()p r < a jp r. By the well ordering of the integers, by the fact that 0 < a, and by the fact that eventually some multiple of p r is a, such a j exists. Since eventually a multiple of p r will be > a + b, there is l N {0} such that (j + l )p r a + b < (j + l)p r. (l = 0 if already jp r > a + b.) To figure out how many multiples of p r lie between a +, a + b (endpoints included) we consider two cases. Case : a = jp r. In this case the multiples are (j + )p r,..., (j + l )p r, a total of l. So k r = l. We have a/p r = j, j + l (a + b)/p r < j + l, thus On the other hand if a < jp r, then one sees that k r = l and k r = l = (j + l ) j = a + b p r a p r. < a p r < j = = a p r ; 4

5 from (j + l )p r < a + b (j + l)p r we get j + l a + b p r < j + l = j + l = a + b p r. Thus once more, as stated. k r = l = (j + l ) () = a + b p r a p r, Step 3. We will also need the following simple result. Let a, b, c be positive integers. Then 2a c + 2b c a c b c a + b c Proof. Let k = a/c, j = b/c. Then k a/c < k +, thus ck a < c(k + ); similarly, cj b < c(j + ). Thus ck + cj a + b < c(k + j + 2), hence k + j a + b < k + j + 2. c This implies that (a + b)/c is k + j or k + j +. We need to have an idea when it will be k + j +, and we see that for these to happen at least one of the following two inequalities must hold: a c k + 2, b c j + 2. Without loss of generality we can assume it is the first one. Thus (a + b)/c = k + j, or (a + b)/c is k + j + and a/c > k + (/2). Now 2k 2a/c < 2k + 2 so that 2a/c is either 2k or 2k +. If it is 2k, then 2a/c < 2k +, hence a/c < k + /2 and (by our choice of a as the one where a/c is larger than integer plus one half in case (a + b)/c > k + j) we also have (a+b)/c = k +j. Similarly b/c is 2j or 2j +, so that b/c 2j. Thus, if 2a/c = 2k, then (a+b)/c = k +j and 2a c + 2b c a c b + b 2k + 2j k j = k + j = a. c c On the other hand, if 2a/c = 2k +, 2a c + 2b c a c b + b 2k + + 2j k j = k + j + a. c c Step 4. Pulling it all together. We have (2m)!(2n)! m!n!(m + n)! (m + )(m + 2)... (2m)(n + )(n + 2) (2n) =. (m + n)! We want to see that if p is any prime dividing the denominator, and if p R is the largest power of p dividing the denominator (that is p R divides the denominator, but p R+ does not), then p R divides the numerator. By the uniqueness of the prime power decomposition, this proves the denominator divides the numerator evenly, and the expression is an integer. So let p be a prime factor of the denominator. Let k r be the number of multiples of p r in the list, 2,..., m + n; by Step, By Step 2, with a = 0, b = m + n R = k r. r=0 k r = m + n p r. Now let (I am sorry for all this notation) let p R, p R be the largest powers of p dividing (m + ) (2m), (n + ) (2n), respectively, then we want to show that R R + R. That will do it. For this, for r =, 2,..., let k r be the number of 5

6 multiples of p r greater than or equal m + and less than or equal 2m, let k r be the number of multiples of p r greater than or equal n + and less than or equal 2n. Since R = k r, R = k r, it suffices to prove that k r + k r k r for all r =, 2, By Step 2, so proving k r + k r k r is equivalent to proving r=0 k r = 2m p r m p r, r=0 k r = 2n p r n p r 2m p r m p r + 2n p r n p r m + n p r. However, this is a consequence of Step 3, with a = m, b = n and c = p r. We are done. While this proof is correct, it is particularly non-elegant. There should be a nicer, more illuminating proof. Look for it! 6

2x 1 7. A linear congruence in modular arithmetic is an equation of the form. Why is the solution a set of integers rather than a unique integer?

2x 1 7. A linear congruence in modular arithmetic is an equation of the form. Why is the solution a set of integers rather than a unique integer? Chapter 3: Theory of Modular Arithmetic 25 SECTION C Solving Linear Congruences By the end of this section you will be able to solve congruence equations determine the number of solutions find the multiplicative