Numbers (Part I) -- Solutions

Transcription

1 Ley College -- For AMATYC SML Mth Competitio Cochig Sessios v.., [/7/00] sme s /6/009 versio, with presettio improvemets Numbers Prt I) -- Solutios. The equtio b c 008 hs solutio i which, b, c re distict eve positive itegers. Fid b c. [008S, ] Sol: Let x, b y, c z. The x y z 5. Sice, 8, 7, 6, 5 5, 6 6, 7, we see tht x, y, z re t most 6. Moreover, sice 6 9 5, the lrgest of x, y, z must be 5 or 6. Sy, x 6, the y z 5 6 5, which is 7 8. So x y z 6, i.e. b c. It is ot hrd to see tht this choice of x, y, z d thus, b, c ) is the oly possibility up to reorderig.. Ech bg to be loded oto ple weighs either, 8, or lb. If the ple is crryig exctly 000 lb of luggge, wht is the lrgest umber of bgs it could be crryig? [008S, 8] Sol: We wt to fid whole umbers x, y, z to mximize x y z while mitiig 6x 9y z 500 sme s x 8y z 000). Oe should use smller bgs wheever possible. More precisely, brigig dow z by would reduce 6 x 9y z, by, which c the be compested for by icresig x by d y by becuse 6 9 ), with x y z icresed. So whe x y z is mximized, z must be 0,, or. Likewise, brigig dow y by while icresig x by hs the effect of keepig 6x 9y z 500 while icresig x y z. Therefore, whe x y z is mximized, y must be 0 or. Sice i 6x 9y z 500, 6 x d 500 re eve, so 9 y z hs to be eve. This implies tht y, z) hs to be 0, 0), 0, ),, ). Amog them, oly the lst oe would mke 500 9y z divisible by 6. This results i x 80, y, z, so x y z 8.. Cll positive iteger biprime if it is the product of exctly two distict primes thus 6 d 5 re biprime, but 9 d re o. If N is the smllest umber such tht N, N, N re ll biprime, fid the lrgest prime fctor of N N ) N ). [008S, 7] Sol: Observe tht biprime cot be multiple of. Hece N hs to be cogruet to modulo i.e. of the form k ), for otherwise t lest oe of the three umbers would be multiple of. It follows tht N is eve. But N is coprime, so N p, with p prime, d thus N 6, 0,,, 6,,. We quickly rule out N 6, through 6, s either N or N would fil to be coprime. Thus N, N, N re,, 5, with N N ) N ) The swer follows.. Let r, s, t be oegtive itegers. For how my such triples r, s, rs t stisfyig the system is it true tht r s t 5? [008S, 6] r st Sol: Tke the differece of the two equtios i the system to get r s ) t s) 0, i.e. s ) r 0. Cse : s. Thus the system sys r t d r s t 5 lso becomes r t. There re 5 possible triples Pge of 6

2 Ley College -- For AMATYC SML Mth Competitio Cochig Sessios v.., [/7/00] sme s /6/009 versio, with presettio improvemets r, s, of this kid. Cse : s, thus r t. The system sys t s ), while t s ) r s t 5 sys t s 5. So we hve, which mes t s 5 s ) 8, therefore t d s form the two roots of the qudrtic s ) 6 equtio x 6x 8 0, i.e. x ) x ) 0. Sice s, it follows tht s, t, so r, s,,,). Cses d combied give 6 triples. 5. The digits to 9 c be seprted ito disjoit sets of digits ech so tht the digits i ech set c be rrged to form -digit perfect squre. Fid the lst two digits of the sum of these three perfect squres. A. 6 B. 9 C. D. 6 E. 7 [008S, E] Sol: Approch the problem by brute force. First, list ll perfect squres formed by three distict o-zero digits by squrig 0,,, etc d rulig out those with repeted digits. The resultig list is 69, 96, 56, 89,, 6, 59, 576, 65, 79, 78, 8, 96. Amog them oly d 6 coti the digit, thus oe of them hs to be icluded. If it s, the the other two perfect squres must be from mog 69, 96, 576, 96, i order to void reusig,, or. For the digit 5 to pper, we re forced to iclude 576, with the remiig digits, 8, 9 uble to form perfect squre, thus does t work. Try 6, with the other two perfect squres thus comig from mog 89, 59, 79, 78. To ccommodte the digit 5, we iclude 59, with the remiig perfect squre thus beig 78. We hve The sequece } is defied by, d )! { 0 00, fid k. k for. If 5 is the lrgest power of 5 tht is fctor of 0 [008S, ] Sol: Divide )! o both sides to get by )!. Try the first few itertios to see if there is y ptter: 0!, i.e. 0!, thus 0 0 d!, i.e.!, thus 5!, i.e. 5!, thus !, i.e. 6!, thus We begi to suspect tht i geerl, which the leds to ) ) ). For exmple, 6 5, 7 6, 8 7 5, etc. If such ptter persists up to, the )! becomes )! ), d thus )!, d so Pge of 6

3 Ley College -- For AMATYC SML Mth Competitio Cochig Sessios v.., [/7/00] sme s /6/009 versio, with presettio improvemets the iductio c be cotiued to uphold the ptter. With the ptter ow estblished, it follows tht ! We re left with fidig how my fctors of 5 the umber 00! cotis. Amog the oe hudred umbers through 00, twety umbers re multiples of 5, i.e. 5, 0, 5, 0,, 95, 00. Amog these, 5, 50, 75, 00 hve the fctor 5 pperig two times. So, ltogether 00! hs 0+= fctors of 5. The swer follows. 7. Tri hs two doze cois, ll dimes d ickels, worth betwee $.7 d $.. Wht is the lest umber of dimes she could hve? [007F, ] Sol: Let there be x dimes d y ickels. Thus x y, while 7 0x 5y. Sice 0 x 5y is divisible by 5, so 75 0x 5y 0, i.e. 5 x y, i.e. 5 x x y). As x y, we get x 8. So the lest possible vlue for x is d thus y.) 8. Replce ech letter of AMATYC with digit 0 through 9 to form six-digit umber ideticl letters re replced by ideticl digits, differet letters re replced by differet digits). If the resultig umber is the lrgest such umber which is perfect squre, fid the sum of its digits tht is, A M A T Y C ) A. B. C. D. 5 E. 6 [007F, E] Sol: We first try 989XXX. Such umbers do ot exceed Use clcultor to get , so if 989XXX is perfect squre, it hs to be t most , which is lredy too smll to be of the form 989XXX. This rules out 989XXX. The sme pproch c be used to rule out 979XXX, 969XXX,, 99XXX. For 99XXX, we do , d thus we compute , which is of the form 99XXX, but does t stisfy the required form AMATYC, while is lredy too smll to be of the form 99XXX. This rules out 99XXX. 99XXX c lso be ruled out this wy. 909XXX is lso ruled, i mer similr to 989XXX. We coclude tht A cot be 9. We the try 898XXX, usig the sme method, d immeditely hit The swer follows. 9. Add y iteger N to the squre of N to produce iteger M. For how my vlues of N is M prime? [007F, ] Sol: N N) N N NN ). Sice this is prime, t lest oe of the followig coditios is met: ) N ) N ) N ) N. Of these, oly ) d ) mke sese. The swer follows. 0. Whe certi proper frctios i simplest terms re dded, the result is i 9 simplest terms: 5 05; i other cses, the result is ot i simplest terms: 5 9 m Assume tht 5 d re positive proper frctios i simplest m terms. For how my such frctios is 5 ot i simplest terms? [007F, 8] m 7m 5 m Sol: Sice 5 d re positive proper frctios i simplest terms, we hve m,,, 7, 8,,, d,,, 5, 8, 0,,, 6, 7, 9, 0. 7m 5 Sice 05 is ot i simplest terms, 7 m 5 is divisible by 5, 7, or. But it is ot divisible by 5 becuse m is ot divisible by 5. Likewise 7 m 5 is ot divisible by Pge of 6

4 Ley College -- For AMATYC SML Mth Competitio Cochig Sessios v.., [/7/00] sme s /6/009 versio, with presettio improvemets 7 either. It follows tht 7 m 5 is divisible by. But 7m 5 6 m ) m ), so this is the sme s syig m is divisible by, i.e. m mod. Amog the eight possible vlues for m,,, 7, re cogruet to mod, while the remiig, 8,, re cogruet to mod. For the twelve possible vlues for,,, 0,, 6, 9 re cogruet to mod, while the remiig, 5, 8,, 7, 0 re cogruet to mod. If m is cogruet to mod, the so is, with 6 combitios. Likewise, if m is cogruet to mod, the so is, with 6 combitios. Together we hve therefore 8 scerios.. Let r, s, d t be oegtive itegers. How my such triples r, s, stisfy the system rs t? [007F, ] r st Sol: Subtrct the secod equtio from the first equtio to get r s ) t s), i.e. s s ) r. Two cses: Cse :, so s r t, d the origil system becomes r t s, which gives r 5, t. Cse :, so s r t r t 0, d the origil system becomes t. We coclude tht there re two such triples. r. If AM/AT.YC, where ech letter represets differet digit, AM/AT is i simplest terms, d A 0, the AT? [007S, 5] Sol: AM YC. So AT is fctor of 00. There re ie fctors of 00, mog AT 00 them the followig hve two digits: 0, 0, 5, 50. But AM hs to be proper AT frctio, therefore AT cot be 0, 0, or 50. The swer follows.. Two djcet fces of rectgulr box hve res 6 d 6. If ll three dimesios re positive itegers, fid the rtio of the lrgest possible volume of the box to the smllest possible volume. [006F, 9] Sol: This mes there re positive itegers, b, c with b 6 d bc 6, so b is commo fctor of 6 d 6. Sice the GCF of 6 d 6 is 9, this mes b is fctor of 9, d so b b) bc) 6)6),, 9. The volume is bc b b. It follows tht 6)6) 6)6) the lrgest d smllest possible vlues re d 9. The swer follows.. I the expressio AM)AT)YC), ech differet letter is replced by differet digit 0 to 9 to form three two-digit umbers. If the product is to be s lrge s possible, wht re the lst two digits of the product? A. 0 B. 0 C. 50 D. 60 E. 90 [006F, B] Sol: Try 95)96)87) 790. We will covice ourselves tht this is the lrgest possible. First of ll, t lest oe of the three umbers hve to be t lest 90, for otherwise the product would be t most 89)89)89) But Y cot be 9, for otherwise the product would be t most 89)89)98) It follows tht A is 9. The Y hs to be 8, for otherwise the product would be t most 98)98)79) We re left with oly 9M)9T)8C), d we oly hve to compre the three products: Pge of 6

5 Ley College -- For AMATYC SML Mth Competitio Cochig Sessios v.., [/7/00] sme s /6/009 versio, with presettio improvemets 95)96)87), 95)97)86), d 96)97)85), of which 95)96)87) 790is the lrgest. b If, fid the smllest possible vlue of b c d, c d if, b, c, d d re ll positive itegers. [006F, 6] b b 0 6b 5 b 5 b) Sol:. So the c d 6 5c d 0c 6d 5c d 5c d) b b 0 coditio tht is equivlet to. But c d c d 0 5 b 0 mes :b : 5. So there is positive iteger u such tht u, b 5 u. Likewise, c v, d 5 v for some positive iteger v. So b c d 8u 8v 8 u v). The swer follows. 6. The yer 006 is the product of exctly three distict primes p, q, d r. How my other yers re lso the product of three distict primes with sum equl to p q r? [006F, ] Sol: , with Thus we wt whole umbers tht re the product of three distict primes whose sum is 78. Note tht 78 is eve, so oe of the three primes must be, with the sum of the other two beig 76. List ll primes smller th 80 s follows, from which we idetify pirs tht sum to 76:,, 5, 7,,, 7, 9,, 9,, 7,,, 7, 5, 59, 6, 67, 7, 7, 79 The swer follows. 7. How my positive itegers less th 000 re reltively prime A to 05? Two itegers re reltively prime if their gretest commo divisor is. [006F, 57] Sol: Let U be the set of ll positive itegers less th 000. Let A, B, C be the subsets of U cotiig those tht re divisible by, 5, 7, respectively. The the problem ims to determie C # U A B C)). This equls # U [# A # B # C # A B) # B C) # C A) # A B C)]. Note tht, for exmple, A B is the subset of U formed by those tht re divisible by 5 5, thus # A B) 66 sice Likewise, e.g. A B C is the subset formed by those tht re divisible by Thus we get 999 [ ] How my -digit umbers whose digits re ll odd re multiples of? [006F, 85] Sol: Let ABCD be such -digit umber, with ech of A, B, C, D beig y of,, 5, 7, 9. Divisibility of is equivlet to the coditio tht the differece betwee the sum A+C d the sum B+D is divisible by. But either sum is t lest d t most 9 9 8, so the differece must be, 0, or. Sice both A+C d B U B Pge 5 of 6

6 Ley College -- For AMATYC SML Mth Competitio Cochig Sessios v.., [/7/00] sme s /6/009 versio, with presettio improvemets B+D re eve, we rule out d, thus A+C=B+D. List ll possible ordered pir from,, 5, 7, 9, grouped ccordig to the sum: Sum = : + Sum = : +, + Sum = 6: +5, 5+, + Sum = 8: +7, 7+, +5, 5+ Sum = 0: +9, 9+, +7, 7+, 5+5 Sum = : +9, 9+, 5+7, 7+5 Sum = : 5+9, 9+5, 7+7 Sum = 6: 7+9, 9+7 Sum = 8: 9+9 Both A+C d B+D hve to be from the sme group. There re thus 5 85 possibilities Fid the tes digit of. [006F, 8] 007 Sol : 006 ) 00 9 ) 0 00 ). By the biomil expsio, we see 0 00 ) ) ) ) Thus 0 00 ) ) 00 ) mod 00. So ) )009) mod 00. The swer follows. Sol : Workig modulo 00, we hve, 9, 7, 8, 5, 6 9, 7 87, 8 6, 9 8, 0 9, 7,,, 69, 5 7, 6, 7 6, 8 89, 9 67, 0,. We thus 007 otice tht every 0 steps the thig repets. It follows tht 7 mod 00, i.e mod 00. The swer follows. 0. I the sequece,,,,,, 5, d for ll,. Fid 006/ 005. [006F, 00] Sol: Divide o both sides by to get, i.e.. This shows tht,,, form rithmetic sequece with commo differece. The first term is, so the term 006 is 005 ) ) 00 ) 00 Pge 6 of 6