Solutions to Problem Set 5

Transcription

1 Massahusetts Institute of Tehnology 6.042J/18.062J, Fall 05: Mathematis for Computer Siene Otober 28 Prof. Albert R. Meyer and Prof. Ronitt Rubinfeld revised Otober 12, 2005, 909 minutes Solutions to Problem Set 5 Problem 1. Suppose that one domino an over exatly two squares on a hessboard, either vertially or horizontally. (a) Can you tile an 8 8 hessboard with 32 dominos? dominos hess board Solution. Yes. Plae 4 vertial dominos in eah olumn. (b) Can you tile an 8 8 hessboard with 31 dominos if opposite orners are removed? Solution. No! Opposing orners are the same olor. Therefore, removing opposite orners leaves an unequal number of white and blak squares. Sine every domino overs one blak square and one white square, no tiling is possible. Copyright 2005, Prof. Albert R. Meyer and Prof. Ronitt Rubinfeld.

2 2 () Now suppose that an assortment of squares are removed from a hessboard. An example is shown below. Given a trunated hessboard, show how to onstrut a bipartite graph G that has a perfet mathing if and only if the hessboard an be tiled with dominos. Solution. Create a vertex for every white square and a vertex for every blak square. Put an edge between squares that share an edge. (This graph is bipartite, sine the oloring of the squares defines a valid 2 oloring of the verties.) If a perfet mathing exists in this graph, then a tiling exists: put a domino over eah pair of mathed verties. On the other hand, if a tiling exists, then a perfet mathing exists: math squares overed by the same domino. (d) Based on this onstrution and Hall s theorem, an you state a neessary and suffiient ondition for a trunated hessboard to be tilable with dominos? Try not to mention graphs or mathings! Solution. A board an be tiled with dominos if and only if every set of white squares is adjaent to at least as many blak squares and vie versa. Problem 2. Prove that gd(ka, kb) = k gd(a, b)for all k > 0. Solution. The smallest positive value of: k (s a + t b) (whih is equal to s(ka) + t(kb) = gd(ka, kb)) must be k times the smallest positive value of: s a + t b (whih is equal to gd(a, b)).

3 3 Problem 3. Suppose that a b (mod n) and n > 0. Prove or disprove the following assertions: (a) a b (mod n) where 0 Solution. The proof is by indution on with the hypothesis that a = 0, then the laim holds, beause 1 1 (mod n). Now suppose that: b (mod n). If a b (mod n) Multiplying both sides by a gives: a +1 ab (mod n) Sine a b (mod n), we an replaed the a on the right side by b: +1 a b +1 (mod n) Therefore, the laim holds by indution. (b) a b (mod n) where a, b, 0 Solution. The laim is false. For example: (mod 3) (1) Problem 4. An inverse of k modulo n > 1 is an integer, k 1, suh that k k 1 1 (mod n). Show that k has an inverse iff gd(k, n) = 1. Hint: We saw how to prove the above when n is prime.

4 4 Solution. If gd(k, n) = 1, then there exist integers x and y suh that kx + yn = 1. Therefore, yn = 1 kx, whih means n (1 kx) and so kx 1 (mod n). Let k 1 be x. Problem 5. Here is a long run of omposite numbers: 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126 Prove that there exist arbitrarily long runs of omposite numbers. Consider numbers a little bigger than n! where n! = n (n 1) Solution. Let k be some natural number suh that 1 < k n. We know k (n! + k) beause k n! and k k. Thus, the numbers n! + 2, n! + 3, n! + 4,..., n! + n must all be omposite. This is a run of n 1 onseutive omposite numbers. Beause we an arbitrarily hoose n, we know arbitrarily long runs of ompisite numbers exist. Problem 6. Take a big number, suh as Sum the digits, where every other one is negated: 3 + ( 7) ) ( ) ( ) ( ) ( + 1 = 11 As it turns out, the original number is a multiple of 11 if and only if this sum is a multiple of 11. (a) Use a result from elsewhere on this problem set to show that 10 k 1 k (mod 11). Solution. We know 10 1 (mod 11). From 2a, we onlude 10 k ( 1) k (mod 11). (b) Using this fat, explain why the proedure above works. Solution. A number in deimal has the form: d k 10 k + d k 1 10 k d d 0

5 5 From the observation above, we know: d k 10 k + d k 1 10 k d d 0 d k ( 1) k + d k 1 ( 1) k d 1 ( 1) 1 + d 0 ( 1) 0 (mod 11) d k d k d 1 + d 0 (mod 11) Note that the above assumes k is even. The ase where k is odd is analogous. Also, the proedure given in the problem may have us reverse all signs. Beause we are only heking for divisibility, this does not matter. Problem 7. Let S k = 1 k + 2 k (p 1) k, where p is an odd prime and k is a positive multiple of p 1. Use Fermat s theorem to prove that S k 1 (mod p). Solution. Fermat s theorem says that x p 1 1 (mod p) when 1 x p 1. Sine k is a multiple of p 1, raising eah side to a suitable power proves that x k 1 (mod p). Thus: 1 k + 2 k (p 1) k } {{ } 1 (mod p) p 1 1 p 1 terms (mod p) (mod p)