n Find all pairs of integers x, y which satisfy the diophantine equation 50x 65y = 75. a i. G n = a i a 1 a 2 a k

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1 Math 00. Treibergs σιι Fial Exam Name: Sample March 7, 0 Sample Fial Questios.. Show usig iductio that 5 (7 ) for every positive iteger.. Prove that! > for all itegers 4. [From miterm for Math 30-, Sept. 6, 009] 3. Prove that for all N, i i(i + ) Fid all pairs of itegers x, y which satisfy the diophatie equatio 50x 65y Let p be prime. The for ay iteger x, x p x is divisible by p. 6. Let a, a,..., a be positive real umbers. The Arithmetic Mea of these umbers is defied to be A a i. i The Geometric Mea of these umbers is defied by Show that for all, G A. 7. Show that S G ( i {a,a,...,a k } {,,...,} a i ). a a a k where the sum is take over all oempty subsets of the set of smallest positive itegers. 8. A kight o a chessboard ca move oe space horizotally (i either directio) ad the two spaces vertically (i either directio) or two spaces horizotally (i either directio) ad the oe space vertically (i either directio). Suppose that we have a ifiite chessboard made up of all squares (m, ) where m ad are oegative itegers. Show that a kight startig at (0, 0) ad travellig withi the chessboard ca reach ay square of the chessboard usig a fiite umber of moves. 9. Determie which amouts of postage ca be formed usig just 4 cet ad 7 cet stamps. 0. Let f be the -th Fiboacci umber. Show that for positive iteger we have f + f + + f f f +.. Let f be the -th Fiboacci umber. Show that for positive iteger we have S f 0 f + f f + f f

2 . Let f be the -th Fiboacci umber. Show that for positive iteger we have ( ) ( ) f+ f Show that the -th derivative f (x) d dx ( xe x ) (x + )e x for all positive itegers. 4. Fid the umber of itegers betwee 000 ad 9999 iclusive that are divisible by 7 or are ot divisible by. 5. Suppose that there are fiite sets A, A,..., A. The the umber of elemets i the uio is give by the iclusio-exclusio formula A A ( ) k+ k f f i <i < <i k A i A i A ik where the secod sum is over all k elemet subsets {i, i,..., i k } {,, 3,..., }. Prove the formula usig iductio. (Aother proof is give i 7.5.) 6. Show that for every positive iteger, (p p p ) is equivalet to p p p where p, p,..., p are propositios. 7. Suppose that you begi with a pile of stoes ad split the pile ito piles of oe stoe each by successively splittig a pile of stoes ito two smaller piles. Each time you split a pile, you multiply the the umber of stoes i each of the smaller piles you form, so that if the smaller piles have r ad s stoes i them, you compute rs. Show that o matter how you split the piles, the sum of the products computed at each step equals to ( ). 8. Let d be as positive iteger. Show that i ay group of d + ot ecessarily cosecutive itegers, there must be two with exactly the same remaider whe divided by d. 9. Suppose that there are people at a party. Show that there must be at least two of them who kow the same umber of people at the party. Questios from Math exam of Ja. 30, A stadard deck of 5 cards cosists of four suits {,,, }. Each suit has 3 differet kids of cards {, 3, 4, 5, 6, 7, 8, 9, 0, J, Q, K, A}. How may two card hads are there such that both cards are of the same suit or of the same kid?. A bag cotais 6 scrabble tiles, each labeled by a differet letters of the alphabet. (a) How may differet five letter words ca be selected from the bag (without replacemet)? (b) How may of these words are i alphabetical order? (c) How may of these cotai at least oe vowel {A, E, I, O, U}?

3 Solutios.. Show usig iductio that 5 (7 ) for every positive iteger. Let P () deote the statemet 5 (7 ). For the iductio, we eed to show P () ad P (k) P (k + ) for all k. For, which is divisible by 5. Hece P () is true. Assume for some positive iteger k, 5 divides 7 k k. The 7 k+ k+ 7 7 k k 5 7 k + (7 k k ) 5 divides the first term ad by the iductio hypothesis, P (k), it divides the secod term as well ad thus the sum. Hece we coclude P (k + ), ad the iductio is complete.. Prove that! > for all itegers 4. [From miterm for Math 30-, Sept. 6, 009] Use iductio o. I the base case 4, the Lhs 4! > 6 4 Rhs. Iductio case. Assume that for ay 4 we have! >. The by the iductio hypothesis ad +, 3. Prove that for all positive itegers, ( + )! ( + )! > ( + ) +. i i(i + ) +. Proof by iductio. I the base case,, the left side is right side is + + hece equality holds. i i(i+). The The iductio step is to prove for every, the statemet for + is true assumig it s true for. But + i i(i + ) ( + )( + ) + i i(i + ) Now use the iductio hypothesis. ( + )( + ) + + ( + ) + ( + )( + ) ( + ) ( + )( + ) + ( + ) Fid all pairs of itegers x, y which satisfy the diophatie equatio 50x 65y 75. () Note that 50 5 ad so d gcd(50, 65) 5. Sice 5 divides the left had side, there is o solutio of () uless 5 divides the right side too, which it does. Dividig by d we get 0x 3y 5, () whose solutios are the solutios of (). Now 0 ad 3 are relatively prime so there are umbers l ad m so that 0l 3m. To fid l ad m, we ru the Euclidea Algorithm

4 Substitutig, we fid (3 0) , so l 4 ad m 3. Multiplyig by 5 we fid Thus oe solutio of the diophatie equatio is x 0 60 ad y Suppose (x, y) is aother solutio of (). The 0(x x 0 ) 3(y y 0 ) Sice 0 ad 3 are relatively prime, 0 (y y 0 ) so there is a iteger t so that y y 0 +0t. The 0(x x 0 ) 30t or x x 0 3t. It follows that other solutios have the form x t y t. But by substitutig (3) ito (), we see that this is a solutio for ay t. possible solutios of (), ad hece of () are give by (3). (3) Therefore all 5. Let p be prime. The for ay iteger x, x p x is divisible by p. If p the x x x(x ) cotais a eve factor sice either x or x is eve. Thus we may assume p is a odd prime, ad x 0, sice for odd p, by replacig x by x yields (x p x) ( x) p ( x) which has the same divisibility by p as x p x. The statemet is true for x 0 ad x. We argue by iductio o x. Assume the statemet is true for ay positive iteger x. Observe that (x + ) p p j0 ( ) p x j j But the biomial coefficiets ( ( p ), p ) (,... p p ) are all divisible by p because if j p the the iteger biomial coefficiet ( ) p p(p ) (p j + ) j j(j ) has a sigle largest prime factor p i the umerator ad o prime factor i the deomiator as large as p. Hece p ( p j). Thus, by the iductio hypothesis, (x + ) p x p + x + (mod p), ad the formula holds also for x + : the iductio is complete. 6. Let a, a,..., a be positive real umbers. The Arithmetic Mea of these umbers is defied to be A a i. i The Geometric Mea of these umbers is defied by G ( 4 i a i ).

5 Show that for all, G A. The trick that makes the argumet easier is to double the umber of terms each iductio step rather tha just icreasig the umber by oe. The, by a simple paddig argumet, the geeral result follows. We argue by iductio that the iequality is true whe k is a power of two. I the base case k 0 or 0, A a G ad so G A. We illustrate the cases ad to make the geeral iductio step clearer. If there are two umbers α ad β, the basic squarig iequality is from which follows. 0 ( α β) α αβ + β G αβ (α + β) A If there are four umbers α, β, γ ad δ, apply the basic squarig iequality twice ( ) G 4 (αβγδ) 4 αβ γδ ( ) αβ + γδ ( α + β + γ + δ ) 4 (α + β + γ + δ) A 4. Now, for ay 0, assume the result is true for terms: G A. Applyig first the squarig iequality ad the the iductio hypothesis to both to the first umbers ad also to the secod umbers ( + + ) G + + a i i a i i ( ) a i i ( + ) a i i j a i A +. i a j + j + a j + j + a j To illustrate the paddig, if there are three umbers, the use A 3 as a fourth ad the the Arithmetic-Geometric iequality for four umbers gives G A 4 3 (αβγa 3 ) 4 4 (α + β + γ + A 3) 4 (3A 3 + A 3 ) A 3 from which G 3 A 3 follows. I geeral, if k < k+, we pad the umbers with k+ copies of A ad use the Arithmetic-Geometric iequality for k+ umbers ( ) G k+ k+ k+ A k+ A k+ a i ( ) k+ ( k+ )A + A A from which G A follows. i 5

6 7. Show that S {a,a,...,a k } {,,...,} a a a k where the sum is take over all oempty subsets of the set of smallest positive itegers. We argue by iductio o. For the base case whe, the oly subset is {} {}, so the sum is / which equals. Assume for ay that the statemet is true for sums over subsets of the first itegers. Now, partitio the sum ito three parts. A subset of {,,...,, +} either is the sigleto set { + }, a subset that cotais + but has also aother elemet, or is a subset that does ot cotai +. I the latter case, the subset may be viewed as a oempty subset of {,,..., }. Thus S + {a,a,...,a k } {,,...,+} {+} {,,...,+} {a,a,...,a k } {,,...,} a a a k {a,a,...,a k,+} {,,...,+} a a a k a a a k ( + ) Notig that there is oe term i the first sum, that all secod sum subsets are of the form of T { + } where T {a, a,..., a k } {,,..., } so that + ca be factored out of the sum, we see that usig the iductio hypothesis o the secod ad third terms S {a,a,...,a k } {,,...} The iductio step is complete. a a a k + {a,a,...,a k } {,,...,} a a a k 8. A kight o a chessboard ca move oe space horizotally (i either directio) ad the two spaces vertically (i either directio) or two spaces horizotally (i either directio) ad the oe space vertically (i either directio). Suppose that we have a ifiite chessboard made up of all squares (m, ) where m ad are oegative itegers. Show that a kight startig at (0, 0) ad travellig withi the chessboard ca reach ay square of the chessboard usig a fiite umber of moves. We argue usig iductio o m + which rus through all oegative itegers for chessboard squares. A kight move cosists of addig oe of eight vectors to the curret positio. These eight are (±, ±) or (±, ±). The kight may ot move off of the board. The statemet to be verified is P(k) Startig from (0, 0), the kight ca reach the square (m, ) where m + k by usig at most l k 3k moves o the chessboard. The base case is k 0. The kight eed ot move to reach the oly square with m + 0, amely (0, 0). Hece l 0 0. For ay k 0, the iductio hypothesis is that we assume that the kight ca reach ay square (m, ) o the diagoal m + k i at most l k 3k moves. Cosider ay square (m, ) where m + k +. Sice the sum is positive, oe of the umbers must be positive. I case m > 0, we kow that m is oegative ad (m, ) is a square o the (m ) + k diagoal. By the iductio hypothesis, it takes at most l k 3k 6

7 moves to reach the square (m, ). It takes at most three more moves to reach (m, ), amely (, ) the (, ) the (, ). The positios of the kight are (m, ) (m, + ) (m +, + ) (m, ). Note that the x-coordiates 0 m m m + are all oegative ad the y- coordiates are also oegative. I case m 0 we have > 0, thus is oegative ad (m, ) is a square o the m + ( ) k diagoal. By the iductio hypothesis, it takes at most l k 3k moves to reach the square (m, ). It takes at most three more moves to reach (m, ), amely (, ) the (, ) the (, ). The positios of the kight are (m, ) (m +, ) (m +, + ) (m, ). Note that the x-coordiates 0 m m + m + are all oegative ad the y- coordiates 0 + are also oegative. I both cases, these moves stay i the board. Also, there were three additioal moves, so the total is at most l k+ l k + 3 3k + 3 3(k + ). The iductio is complete. The umber of moves to reach a square may well be less, for example it takes oe move to reach (, ) where + 3 but l A more complicated iductio may yield a sharper estimate. 9. Determie which amouts of postage ca be formed usig just 4 cet ad 7 cet stamps. We solved the Diophatie Equatio before, ad checked whether ay solutios are pairs of oegative itegers. This time we use strog iductio. The first step is to try some sums ad guess which values are possible. We get multiples of four: 0, 4, 8,, 6, 0, 4, 8... ; with oe seve cet stamp: 7,, 5, 9, 3, 7... ; with two seve cet stamps; 4, 8,, 8,... ; with three seve cet stamps:, 8,... ; with four seve cet stamps: 8,.... It is plausible to guess that the amouts of postage possible are 0, 4, 7, 8,,, 4, 5, 6 ad all k 8. Let us prove the result usig strog iductio. The smaller values occur i the table, so are possible. It remais to prove for k 8 the propositio P(k) k cets postage ca be formed usig just 4 cet ad 7 cet stamps. The base cases are the four statemets P(8), P(9), P(0) ad P(). The oegative solutio of 4x + 7y k for k 8 is (, ); for k 9 is (3, ); for k 0 is (5, 0); ad for k is (0, 3). The iductio hypothesis for ay k is P(k 3) P(k ) P(k ) P(k). We show that this implies that P(k +) is true. Usig the assumptio P(k 3), we make postage for k 3 usig (x, y) four ad seve cet stamps, so that 4x + 7y k 3. Addig oe more four cet stamp gives (x +, y) stamps which total 4(x + ) + 7y (4x + 7y) + 4 (k 3) + 4 k + cets postage. Thus P(k + ) is also true, completig the iductio step. 0. Let f be the -th Fiboacci umber. Show that for positive iteger we have The Fiboacci umbers are defied by f + f + + f f f +. f 0 0, f, f f + f for, 3, 4,.... Thus f, f 3, f 4 3, f 5 5, f 6 8, ad so o. 7

8 Let s argue by iductio. For the base case, the left side is f ad the right side is f f so the equatio holds for. For ay, assume that the formula holds for (iductio hypothesis). The for +, by the iductio hypothesis ( + ) fi + f+ f f + + f+ (f + f + )f +. i i f i However, the Fiboacci umbers satisfy f + f + f + so this yields + fi i so the iductio step is verified. (f + f + )f + f + f + f + f (+)+. Let f be the -th Fiboacci umber. Show that for positive iteger we have S f 0 f + f f + f f Let s argue by iductio. For the base case, the left side is f 0 f +f ad the right side is f 0 so the equatio holds for. For ay, assume that the formula holds for. The for +, by the iductio hypothesis S + f 0 f + f f + f f + + f + (f 0 f + f f + f ) f + + f + (f ) f + + f +. However, the Fiboacci umbers satisfy f + f f + ad f + f + f + so these yield S + (f + f ) f + + (f + f + ) f + f (+). so the iductio step is verified.. Let f be the -th Fiboacci umber. Show that for positive iteger we have ( ) ( ) A f+ f 0 ( ) ( ) Let s argue by iductio. For the base case, the left side is ad ( ) ( ) 0 0 f f the right side is so the equatio holds for. f f 0 0 Assume for ay that the formula holds for. The for +, by the iductio hypothesis ( ) + ( )( ) ( )( ) f+ f 0 f f ( ) f+ + f f + f. f + f f f 8

9 However, the Fiboacci umbers satisfy f + + f f + ad f + f f + so these yield ( ) + 0 so the iductio step is verified. ( ) ( f+ f + f(+)+ f + f + f f + f (+) 3. Show that the -th derivative f (x) d dx ( xe x ) (x + )e x for all positive itegers. We ll use iductio ad the product rule. Note that f is iductively defied: f 0 (x) xe x ad f (x) (f (x)) for a positive iteger. The base case is. f (x) (xe x ) x e x + x(e x ) e x + xe x (x + )e x. Now assume for ay that the formula holds for f (x). defiitio, iductio hypothesis ad product rule, ). The, by the recursive f + (x) (f (x)) ((x + )e x ) (x + ) e x +(x+) (e x ) e x +(x+)e x (x++)e x so the formula holds for f + (x) too ad the iductio is complete. 4. Fid the umber of itegers betwee 000 ad 9999 iclusive that are divisible by 7 or are ot divisible by. Let U, the uiversal set, be umbers betwee 000 ad 9999 iclusive. Let A deote the umbers i this rage divisible by seve. Let A deote the umber divisible by eleve. Note that the umbers divisible by seve or ot divisible by eleve may be writte A A (A A ) A. But sice A A ad A are disjoit, by the sum formula A A A A + A. Let us cout the umber both divisible by seve ad divisible by eleve, (amely those divisible by 77). Thus we cout the umber of iteger solutios of the iequality t Thus /77 t 9999/ , hece there are A A solutios. The to fid the umber divisible by eleve, we cout the umber of iteger solutios of the iequality 000 t Because / t 9999/ 909, there are A solutios. The umber of itegers from 000 to 9999 iclusive is U Thus the umber ot divisible by eleve is A Thus, substitutig i the sum formula A A The solutio may also be obtaied usig the iclusio-exclusio formula. 5. Suppose that there are fiite sets A, A,..., A. The the umber of elemets i the uio is give by the iclusio-exclusio formula A A ( ) k+ k i <i < <i k A i A i A ik where the secod sum is over all k elemet subsets {i, i,..., i k } {,, 3,..., }. Prove the formula usig iductio. 9

10 Let us use strog iductio. For oe set, the left side is A ad the right side is ( ) + A, which are equal, so that the base case is verified. For two sets, we ca write the uio as the uio of three disjoit sets, amely A A (A A ) (A A ) (A A ). We ca write each set as a disjoit uio too, A (A A ) (A A ); A (A A ) (A A ). Usig the sum formula, that we ca add the umbers of disjoit sets A A A A + A A + A A which is the case. ( A A + A A ) + ( A A + A A ) A A A + A A A ( ) + ( A + A ) + ( ) + A A For the iductio step, assume that the for some we kow that the formula holds for ay umber k of sets, where k. The, to the uio of + sets, apply the uio formula for the case first A A A + (A A ) A + A A + A + (A A ) A + A A + A + (A A + ) (A A + ) where we have used the distributive law for itersectios ad uios. Now we apply the iductio hypothesis usig the set formula twice A A A + ( ) k+ A i A i A ik + A + ( ) k+ k ( ) k+ + k ( ) k+ k k i <i < <i k i <i < <i k i <i < <i k i <i < <i k (A i A + ) (A i A + ) (A ik A + ) A i A i A ik + ( ) + A + A i A i A ik A +. It remais to regroup the terms to recogize the desired right side. Note that the first sum is over all subsets of {,,..., + } ot ivolvig +, the secod is the sigleto { + } ad the last is over subsets of two or more elemets, oe of which is +. This is the same partitio we ecoutered i problem (7). Note that i each sum the power of is oe more tha the umber of sets i the itersectio. Combiig terms with like umber of sets we complete the iductio + A A A + ( ) k+ k i <i < <i k + A i A i A ik. 0

11 6. Show that for every positive iteger, (p p p ) is equivalet to p p p where p, p,..., p are propositios. Let us use strog mathematical iductio. For the base case, the left side is (p ) ad the right side is p which are logically equivalet sice they are the same statemet. For the case, the statemet is just De Morga s Law of Table 6 o p. 4, which is proved usig a truth table o p.. Thus we have (p p ) p p. Let us ow assume that the statemets are equivalet for some umber of propositios. The for + propositios, usig the equivalece first, ad the the iductive propositio equivalece, (p p p p + ) ([ p p p ] p+ ) ad the iductio step is proved. [ p p p ] p+ [ p p p ] p + p p p p + Note that we have implicitly used the associative property for cojuctio ad disjuctio of arbitrarily may propositios. 7. Suppose that you begi with a pile of stoes ad split the pile ito piles of oe stoe each by successively splittig a pile of stoes ito two smaller piles. Each time you split a pile, you multiply the the umber of stoes i each of the smaller piles you form, so that if the smaller piles have r ad s stoes i them, you compute rs. Show that o matter how you split the piles, the sum of the products computed at each step equals to ( ). For example, if the origial pile has five stoes the, the sequece of splittigs results i 4 4, 4,, whose sum is For aother sequece of splittigs results i 3 6,,, whose sum is also. Let us use strog iductio. For oe stoe, o splittigs are doe so the sum of products is zero whereas 0 0 so the umbers are equal. Now assume for ay positive, the splittig formula holds for ay umber k of stoes, where k. Startig with a pile of + stoes, assume that a arbitrary first splittig divides the pile ito two smaller piles cotaiig r a s stoes, where r + s +. This first splittig cotributes rs to the sum. Note that both r, s because each small pile is oempty ad therefore, both are smaller tha the origial pile. Note that further splittigs of the whole pile either split the first smaller pile or split the secod smaller pile ad whichever pile they split, there is o effect o the splittigs of the other pile, so may be regarded as splittigs of the smaller piles separately. Applyig the iductive hypothesis, the further splittigs of the first pile cotributes r(r ) ad the splittigs of the secod cotributes s(s ) to the sum. Addig the three rs + r(r ) + s(s ) so the iductio is complete. rs + r r + s s (r + s)(r + s ) ( + )

12 8. Let d be as positive iteger. Show that i ay group of d + ot ecessarily cosecutive itegers, there must be two with exactly the same remaider whe divided by d. This is a example of the pigeo hole priciple. d + pigeos (umbers) must reside i d holes (cogruece classes modulo d) ad therefore at least two umbers must belog to the same residue class. 9. Suppose that there are people at a party. Show that there must be at least two of them who kow the same umber of people at the party. We assume that kowig is a symmetric relatio: if A kows B the B kows A. Let v(i) be the umber of other people that the ith perso kows. The set of umbers kow is S {v(i), v(),..., v()}. People may kow ooe or may kow everyoe else at the party, so that 0 v(i). If ooe kows everybody at the party the v(i) < so the fuctio takes v : {,,..., } {0,,..., }. Sice the map is from a larger set to a smaller oe, by the pigeo-hole priciple there must be two people i j such that v(i) v(j). O the other had, there may be sombody, say perso i 0, who kows everybody, v(i 0 ). But the, by symmetry, everyoe kows perso i 0, so v(j) for all j. This time the fuctio maps v : {,,..., } {,,..., }. Sice the map is from a larger set to a smaller oe agai, by the pigeo-hole priciple, there must be two people i j such that v(i) v(j). I either case, we have show that there are at least two people who kow the same umber. 0. A stadard deck of 5 cards cosists of four suits {,,, }. Each suit has 3 differet kids of cards {, 3, 4, 5, 6, 7, 8, 9, 0, J, Q, K, A}. How may two card hads are there such that both cards are of the same suit or of the same kid? We observe that if the had has cards of the same kid they caot be of the same suit ad if the cards have the same suit they caot be of the same kid. Therefore the umber of hads is the sum of hads of the same suit plus hads of the same kid. The umber of hads of the same suit is the umber of suits, 4 times the umber of two card combiatios chose form the thirtee cards of that suit, ( ) 3. The umber of hads of the same kid is umber of kids 3 times the umber of two card combiatios of the same kid, ( 4 total is thus 4 ( ) ( ) A bag cotais 6 scrabble tiles, each labeled by a differet letters of the alphabet. ). The (a) How may differet five letter words ca be selected from the bag (without replacemet)? (b) How may of these words are i alphabetical order? (c) How may of these cotai at least oe vowel {A, E, I, O, U}? (a.) I choosig all words, order is importat. Thus we cout the umber of permutatios of 6 letters, take 5 at a time. The umber of five letter words is P (6, 5) (b.) Give five differet letters, there is oly oe permutatio of these i alphabetical order. Thus the umber of words i alphabetical order is the umber of combiatios of 6 letters take five at a time C(6, 5) ! (c.) The umber with at least oe vowel is the total umber mius the umber with o vowel at all. There are o-vowels so the umber of words without vowels is the umber of permutatios of letters take five at a time P (6, 5) P (, 5)