# graded hw 7 - Math 300 Homework 7 Due Monday 14 November...

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Unformatted text preview: Math 300, Homework 7 Due Monday 14 November, 2011 /(1) Find d = gcd(1071, 462) and ﬁnd integers s and t such that d = 1071.3 + 46223. V(2) Find d = gcd(l20, 23) and ﬁnd integers s and t such that d = 1208 + 23t. /(3) Let a, b E N and let p E P, that is, p is prime. Show that gcd(p, a) = 1 if and only ifp does not divide a. .t ‘/(4) Let a, b E N. Suppose gcd(a, b) = 1. Show that for each n E N, gcd(na, nb) = n. ‘/(5) Let a, b, m E Z, where gcd(a, m) = 1 and m|ab. Show that mlb. (6) Let a, b E Z, not both zero. Show that gcd(a, b) = gcd(a, b + ax) for any x e Z. (7) Let a E b(mod m) and c E d(mod Show that ,‘ (i)a—CEb—d(modm). “E? (ii) ac E bd(mod (8) This question involves solving various congruence equations. (a) If a E imodl then ﬁnd all b E Z such that b E 2a mod 7. (b) If a 2i mod 3 then ﬁnd all b e Z such that b 5 5a4 mod 3. (c) Find all a: E Z such that 2(3: —— 1) E 0 mod 4. (d) Does the congruence equation 0:2 E 0 mod 16 have any solutions x E Z? "w-sgvy , , _ _ NH .1. ot=9c¢00ﬂﬂu©7 / 2}“#213119”54(9753007!16%)) iOH = Maw“ a; , T2497: 300m * (gum) = mm.) @007”) 11491 = sum; 1,! ‘gdmmwwm . .. “M” S J: , m mmo SLgmﬁkngﬂAmema) + ganja). a2. d‘ﬂchILO,L?2):LPrLl§)‘% J ' ’10=.5L1’é) * 5 5: \(2)+l“‘3“’ WNW €5‘Z)*K® =23«J+L5) 4(2): Er- 4(5) + 2, 2= who _ 13-g{.2owa)-»L2)= 2,,,)—|2o(4 4(2) 5: 1(3) + 2 __ gamma/23 ~1 . zmeino-saahﬁ‘nuw; , 3.pmpzs‘)tlonz mom 6N 3mm 126?, ﬁrm in; ,. p is. 9mm. H161: gmtmak l. x? 3m} (mm m a. ; 1 £ 3 . 17m: 'ASSUML gammah. La: asi'xuf 2m) A ’X a. :me 3= ) . . Tmem smuvtp , Visa ommm 'p mum 1W1 gmuﬁfﬁw? - Assam mar p‘ta ,bv’r 9onL‘p,&) >me WW dt‘hnmmﬁF a @012 {.0 nwmr ,qu \ amp dMan 1).me ma \sWakwkdm. 4 .mmmm: Lat maﬁm gamma - 3mm» R1! Bach we M m augmmmm. \$96 We :- L£+ mam ex svgmna& gmwmhrmxg mm mm a and b A no .Eq Wu Wa‘mm Mam-moi Awmwm‘xc, a. bommm mow pv'wm mmwms ,. a=mz...?& , ,XL'EW ml . 1mm . bzdj.‘«5’l 11‘ lgjejP-AQﬂ: me-» \84"j’2,‘!. Kn) L“? m: L15 V4,.L.1V39x%§o& ﬁvaxaqnbYVL .ésmmmn: L2}: 3.,b,vv\€ik .nmm gmcamvu gm mm) ﬁm Max mm, pm: 31002. ggdCeﬁwa.=1 ,5; am m Mam; (“330m mm dummgzmw Lek .a.=«‘.m;mn_ mwnjhng ., 9K1,'433€F W 1;? «.46 Aeqdn. km 1mm dicmmmm QF @ : 2L2; .,._.2.V.\. ms, m pnm dnmﬁm 09 49"Xfxzw’lullrlzw-ﬁn- SW mlab rm: must m 1dr” Xlgfkﬂéz “EL 31M 53:; . mm smelewMMK \S m mm Wm M1,... n W 15w.,ru3k,‘mmm max 396. an mka WW \$31”;de gagging; Tm mm mm Jr‘w m 119, f O 3coz1>aomsﬁim -. Lat we at mam“ m0 mm amt (a.b)=gcc& (a, tram) {Def am "Lé‘ ,. V ., 22/ w W geotcam) “Win via am. vI=> 3Mkamramhw, _ Nu «4 - :3qu (3., W9fo mun we _. Ire/x .. Wmm‘ore lﬂmmm =b. and, waged Lamb =1 mg NM @215 and «3 :1. WM gm (94a): cdge.!0.+9’x) " _ H1) 3?- bmogkm Mina-b). (SI-b.7- mK KJLU’E - Qf— ggmdgm) Mtg/Q) 0d: MIL :LCQ‘EDEMMWJ <2 3—: - 00%) = MLYCX) mm m a2 bLWDgt/m)‘ aQ—bak= Km : 6Q-<§0\,_+80\ -loa\ 4014296., S‘b‘vm Mei cs d—Umqtmlj ‘- aLc-QL) F0t (am) Q‘dFlm ' ‘ ' age. mmoaxmi = 3 {km} (1 (pm)= mCan d1?) . SO, avbovm W K: 31%;). 4X._a.\\$as mejr’mnﬁman be} swnfnaxr DE Za . 6:2qu => zaE‘Kmodﬂr 'bEZa wmué» 235.0%le __ ,.. ., _ mm“ b3 ‘1. 02;. .K£%-b=1tk+<é. L\ 74046 m. 7. g b. was) m5 1m ﬁmau bﬁi mm b=5é4wm5 a: I mow L=> a“ e wvmi © Leg”:— wmm ) *5 41> .63.“: 5m3 b.3561m0‘aLZzMQ’ebé-EW b~5=3x 1 Mi ibai 32K. 'L (LEM a“ me: Smﬂm ﬂax—0:6)le 2(x—n)50mm4© 4 1. le-D-O é; Maw) 4:» ml. 1-! I 2k- M 2M =m . MS 21m 1MB. “Um; Mst whan Maia Mir-Tm W£©X<=5 5:553 ,2..3,5.A,§ 0L mm am Wit) 007—. m \de r” , Moe-o WW Kez xwx Meiqﬂﬂeig RS}: «:0 9&0 Musk-Hg ﬁmJ ‘ “x . WW] ILollwl/ ’xri? \kqlto4=4\/ M2 \uMA=:{/ ...
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graded hw 7 - Math 300 Homework 7 Due Monday 14 November...

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