Cac ham so so hoc

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<ul><li><p>i Hc Thi NguynTrng i Hc Khoa Hc</p><p> Cao Sn</p><p>CC HM S HC V NG DNG</p><p>Chuyn ngnh: PHNG PHP TON S CP</p><p>M S: 60.46.40</p><p>LUN VN THC S TON HC</p><p>Ngi hng dn khoa hc: GS.TSKH. H HUY KHOI</p><p>Thi Nguyn - 2011</p></li><li><p>Cng trnh c hon thnh tiTrng i Hc Khoa Hc - i Hc Thi Nguyn</p><p>Ngi hng dn khoa hc: GS.TSKH. H HUY KHOI</p><p>Phn bin 1: PGS.TS. L Th Thanh Nhn</p><p>Phn bin 2: TS. Nguyn Vn Ngc</p><p>Lun vn c bo v trc hi ng chm lun vn hp ti:Trng i Hc Khoa Hc - i Hc Thi Nguyn</p><p>Ngy 09 thng 09 nm 2011</p><p>C th tm hiu tiTh Vin i Hc Thi Nguyn</p></li><li><p>1Mc lc</p><p>Mc lc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1</p><p>M u . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3</p><p>1 Cc hm s hc c bn 5</p><p>1.1. Phi - hm -le . . . . . . . . . . . . . . . . . . . . . . . 5</p><p>1.1.1. nh ngha . . . . . . . . . . . . . . . . . . . . . 5</p><p>1.1.2. Cc tnh cht . . . . . . . . . . . . . . . . . . . . 6</p><p>1.2. Hm tng cc c s dng ca n . . . . . . . . . . . . . 9</p><p>1.2.1. nh ngha . . . . . . . . . . . . . . . . . . . . . 9</p><p>1.2.2. Cc tnh cht . . . . . . . . . . . . . . . . . . . . 10</p><p>1.3. Hm tng cc ch s ca s t nhin n . . . . . . . . . . 12</p><p>1.3.1. nh ngha . . . . . . . . . . . . . . . . . . . . . 12</p><p>1.3.2. Cc tnh cht . . . . . . . . . . . . . . . . . . . . 12</p><p>1.4. Hm s cc c (n) . . . . . . . . . . . . . . . . . . . . 15</p><p>1.4.1. nh ngha . . . . . . . . . . . . . . . . . . . . . 15</p><p>1.4.2. Cc tnh cht . . . . . . . . . . . . . . . . . . . . 15</p><p>1.5. Hm phn nguyn [x] . . . . . . . . . . . . . . . . . . . . 16</p><p>1.5.1. nh ngha . . . . . . . . . . . . . . . . . . . . . 16</p><p>1.5.2. Cc tnh cht . . . . . . . . . . . . . . . . . . . . 16</p><p>2 Mt s ng dng ca cc hm s hc 18</p><p>2.1. ng dng ca Phi - hm -le . . . . . . . . . . . . . . . 18</p><p>2.1.1. Xt ng d mul ca mt s nguyn t . . . . 18</p><p>2.1.2. Chng minh php chia vi d . . . . . . . . . . . 19</p><p>2.1.3. Gii phng trnh ng d . . . . . . . . . . . . . 20</p><p>2.1.4. Tm nghim nguyn ca phng trnh . . . . . . 21</p></li><li><p>22.1.5. Tm cp ca s nguyn . . . . . . . . . . . . . . . 22</p><p>2.1.6. Tm s t nhin tha mn tnh cht hm s (n) 23</p><p>2.2. ng dng ca hm tng cc c s dng ca s t nhin n 24</p><p>2.2.1. Chng minh mt s l hp s . . . . . . . . . . . 24</p><p>2.2.2. Chng minh mt s l s hon ho . . . . . . . . 25</p><p>2.2.3. Chng minh bt ng thc lin quan ti (n) . . 29</p><p>2.3. ng dng ca hm S(n) . . . . . . . . . . . . . . . . . . 32</p><p>2.3.1. Tm n bi S(n) tha mn mt h thc cho trc . 32</p><p>2.3.2. Tnh gi tr S(n) . . . . . . . . . . . . . . . . . . 35</p><p>2.3.3. Chng minh mt s biu thc lin quan ti S(n) . 37</p><p>2.3.4. Xt tnh b chn ca hm s cha S(n) . . . . . . 39</p><p>2.4. ng dng ca hm s cc c (n) . . . . . . . . . . . . 40</p><p>2.4.1. Tm n tha mn mt iu kin cho trc ca (n) 40</p><p>2.4.2. Mt s bt ng thc lin quan ti hm (n) . . 43</p><p>2.4.3. Tm s nghim ca phng trnh bng phng</p><p>php s dng (n) . . . . . . . . . . . . . . . . . 45</p><p>2.5. ng dng ca hm phn nguyn [x] . . . . . . . . . . . . 46</p><p>2.5.1. Bi ton nh tnh . . . . . . . . . . . . . . . . . 46</p><p>2.5.2. Bi ton nh lng . . . . . . . . . . . . . . . . . 50</p><p>Kt lun . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54</p><p>Ti liu tham kho . . . . . . . . . . . . . . . . . . . . . . . 55</p></li><li><p>3M u</p><p>S hc l mt trong nhng lnh vc c xa nht ca Ton hc, v</p><p>cng l lnh vc tn ti nhiu nht nhng bi ton, nhng gi thuyt</p><p>cha c cu tr li. Trn con ng tm kim li gii cho nhng gi</p><p>thuyt , c nhiu t tng ln, nhiu l thuyt ln ca ton hc </p><p>ny sinh. Hn na, trong nhng nm gn y, S hc khng ch l mt</p><p>lnh vc ca ton hc l thuyt, m cn l lnh vc c nhiu ng dng,</p><p>c bit trong lnh vc bo mt thng tin. V th, vic trang b nhng</p><p>kin thc c bn v s hc ngay t trng ph thng l ht sc cn</p><p>thit. Khng nh nhiu ngnh khc ca ton hc, c rt nhiu thnh</p><p>tu hin i v quan trng ca S hc c th hiu c ch vi nhng</p><p>kin thc ph thng c nng cao mt bc. Do , y chnh l lnh</p><p>vc thun li a hc sinh tip cn nhanh vi khoa hc hin i. Tuy</p><p>nhin, trong chng trnh S hc trng ph thng hin nay, mn S</p><p>hc cha c ginh nhiu thi gian. Cng v th m hc sinh thng</p><p>rt lng tng khi gii bi ton S hc, c bit l trong cc k thi chn</p><p>hc sinh gii.</p><p>Trong phn S hc, cc hm s hc ng vai tr quan trng trong</p><p>vic hnh thnh v nghin cu l thuyt hon thin. y l mt vn</p><p> c in v quan trng ca S hc. Cc bi tp ng dng cc hm s</p><p>hc c bn c cp nhiu trong cc k thi chn hc sinh gii cp</p><p>tnh (thnh ph), Quc gia, Quc t.</p><p>Mc ch chnh ca lun vn l nu ra c mt s ng dng c bn</p><p>ca cc hm s hc c bn (Phi-hm -le, hm tng cc c dng ca</p><p>n, s cc c dng ca n, tng cc ch s ca s t nhin n, hm phn</p><p>nguyn). C th l phn loi c cc dng bi tp ca cc hm s hc</p><p>thng qua h thng bi tp s dng cc hm s hc v cc nh l c</p></li><li><p>4bn ca S hc.</p><p>Ni dung ca lun vn gm 2 chng</p><p>Chng 1: Trnh by cc kin thc c bn ca cc hm s hc.</p><p>Chng 2: Mt s ng dng ca cc hm s hc.</p><p>Lun vn ny c hon thnh vi s hng dn v ch bo tn tnh</p><p>ca GS.TSKH. H Huy Khoi - Vin Ton Hc H Ni. Thy dnh</p><p>nhiu thi gian hng dn v gii p cc thc mc ca ti trong sut</p><p>qu trnh lm lun vn. Ti xin c by t lng bit n su sc n</p><p>Thy.</p><p>Ti xin cm n ti S Ni V, S Gio dc v o to tnh Bc Ninh,</p><p>trng THPT Thun Thnh 1, t Ton trng THPT Thun Thnh 1</p><p> to iu kin gip ti hon thnh kha hc ny.</p><p>Ti xin gi ti cc Thy C khoa Ton, phng o to sau i hc</p><p>Trng i Hc Khoa Hc - i Hc Thi Nguyn, cng nh cc Thy</p><p>c tham gia ging dy kha Cao hc 2009-2011 li cm n su sc v</p><p>cng lao dy d trong sut qu trnh gio dc, o to ca nh trng.</p><p>ng thi ti xin gi li cm n ti tp th lp Cao Hc Ton K3A</p><p>Trng i Hc Khoa Hc ng vin gip ti trong qu trnh hc</p><p>tp v lm lun vn ny.</p><p>Tuy nhin do s hiu bit ca bn thn v khun kh ca lun vn</p><p>thc s, nn chc rng trong qu trnh nghin cu khng trnh khi</p><p>nhng thiu st, ti rt mong c s ng gp kin ca cc Thy C</p><p>v c gi quan tm ti lun vn ny.</p><p>Thi Nguyn, ngy 31 thng 07 nm 2011</p><p>Tc gi</p><p> Cao Sn</p></li><li><p>5Chng 1</p><p>Cc hm s hc c bn</p><p>1.1. Phi - hm -le</p><p>1.1.1. nh ngha</p><p>nh ngha 1.1. Gi s n l mt s nguyn dng. Phi-hm -le ca</p><p>n l s cc s nguyn dng khng vt qu n v nguyn t cng nhau</p><p>vi n.</p><p>K hiu Phi-hm -le l (n).</p><p>V d 1.1. (1) = 1, (2) = 1, (3) = 2, (4) = 2, (5) = 4.</p><p>nh ngha 1.2. Cho n l s nguyn dng. Nu a l s nguyn vi</p><p>(a, n) = 1 th lun tn ti s nguyn dng k ak 1(mod n).S nguyn dng k b nht tha mn ak 1(mod n) c gi lcp ca s nguyn a (modn).</p><p>nh ngha 1.3. Mt h thng d thu gn mul n l mt tp hp</p><p>gm (n) s nguyn sao cho mi phn t ca tp hp u nguyn t</p><p>cng nhau vi n v khng c hai phn t khc nhau no ng d mul</p><p>n.</p><p>V d 1.2. Tp hp {1, 3, 5, 7} l mt h thng d thu gn mul 8.Tp hp {3,1, 1, 3} cng vy.nh ngha 1.4. Mt tp hp A no c gi l mt h thng d y</p><p> (mod n) nu vi bt k s x Z tn ti mt a A x a(modn).</p></li><li><p>6V d 1.3. A = {0, 1, 2, ..., n 1} l mt h thng d y theomul n.</p><p>Ch 1.1. D thy mt tp A = {a1, a2, ..., an} gm n s s l mt hthng d y theo mul n khi v ch khi ai = aj(modn) (ta k hiu"khng ng d" l =) vi i 6= j v i, j {1, 2, ..., n}.</p><p>1.1.2. Cc tnh cht</p><p>Tnh cht 1 . Gi s{r1, r2, ..., r(n)</p><p>}l mt h thng d thu gn mul</p><p>n, a l s nguyn dng v (a, n) = 1. Khi , tp hp{ar1, ar2, ..., ar(n)</p><p>}cng l h thng d thu gn mul n.</p><p>Chng minh. Trc tin ta chng t rng, mi s nguyn arj l nguyn</p><p>t cng nhau vi n. Gi s ngc li, (arj, n) &gt; 1 vi j no . Khi </p><p>tn ti c nguyn t p ca (arj, n). Do , hoc p |a , hoc p |rj , tcl hoc p |a v p |n , hoc p |rj v p |n . Tuy nhin, khng th c p |rj vp |n v rj v n l nguyn t cng nhau. Tng t, khng th c p |a vp |n . Vy, arj v n nguyn t cng nhau vi mi j = 1, 2, ..., (n).</p><p>Cn phi chng t hai s arj, ark (j 6= k) ty khng ng d muln. Gi s arj ark(mod n), j 6= k v 1 j (n) ; 1 k (n). V(a, n) = 1 nn ta suy ra rj rk(mod n). iu ny mu thun v rj, rkcng thuc mt h thng d thu gn ban u mul n.</p><p>V d 1.4. Tp hp {1, 3, 5, 7} l mt h thng d thu gn mul 8.Do (3, 8) = 1 nn {3, 9, 15, 21} cng l mt h thng d mul 8.Tnh cht 2.(nh l -le) Gi s m l s nguyn dng v a l s</p><p>nguyn vi (a,m) = 1. Khi a(m) 1 (modm).Chng minh. Gi s</p><p>{r1, r2, ..., r(n)</p><p>}l mt h thng thu gn gm</p><p>cc s nguyn dng khng vt qu m v nguyn t cng nhau vi m.</p><p>Do Tnh cht 1 v do (a,m) = 1, tp hp{ar1, ar2, ..., ar(n)</p><p>}cng l</p><p>mt h thng d thu gn mul m. Nh vy, cc thng d dng b</p><p>nht ca ar1, ar2, ..., ar(m) phi l cc s nguyn r1, r2, ..., r(m) xp</p><p>theo th t no . V th, nu ta nhn cc v t trong h thng d thu</p><p>gn trn y, ta c: ar1.ar2...ar(m) r1.r2...r(m)(modm).</p></li><li><p>7Do , a(m)r1r2...r(m) r1r2...r(m) (modm).V(r1, r2, ...r(m),m</p><p>)= 1 nn a(m) 1 (modm).</p><p>Ta c th tm nghch o mul n bng cch s dng nh l -le. Gi</p><p>s a,m l cc s nguyn t cng nhau, khi :</p><p>a.a(m)1 = a(m) 1 (modm).Vy a(m)1 l nghch o ca a mul m.</p><p>V d 1.5. 2(9)1 = 261 = 25 = 32 5 (mod 9) l mt nghch o ca2 mul 9.</p><p>H qu 1.1. (a, b) = 1 th a(b) + b(a) 1(mod ab).H qu 1.2. Vi (a, b) = 1 v n, v l hai s nguyn dng no th</p><p>an(b) + bv(a) 1 (mod ab).H qu 1.3. Gi s c k (k 2) s nguyn dng m1,m2, ...,mk vchng nguyn t vi nhau tng i mt. t M = m1.m2...mk = mi.tivi i = 1, 2, ..., k ta c:</p><p>tn1 + tn2 + ...+ t</p><p>nk (t1 + t2 + ...+ tk)n(modM) vi n nguyn dng.</p><p>By gi ta s cho cng thc tnh gi tr ca phi-hm -le</p><p>ti n khi bit phn tch ca n ra tha s nguyn t.</p><p>Tnh cht 3. Vi s nguyn t p ta c (p) = p 1. Ngc li, nu pl s nguyn dng sao cho (p) = p 1 th p l s nguyn t.Chng minh. Nu p l s nguyn t th vi mi s nguyn dng nh</p><p>hn p u nguyn t cng nhau vi p. Do c p 1 s nguyn dng nhvy nn (p) = p 1.</p><p>Ngc li, nu p l hp s th p c cc c d, 1 &lt; d &lt; p. Tt nhin</p><p>p v d khng nguyn t cng nhau. Nh vy, trong cc s 1, 2, ..., p 1phi c nhng s khng nguyn t cng nhau vi p, nn (p) p 2.Theo gi thit, (p) = p 1. Vy p l s nguyn t.Tnh cht 4. Gi s p l s nguyn t v a l s nguyn dng. Khi :</p><p> (pa) = pa pa1.</p></li><li><p>8Chng minh. Cc s nguyn dng nh hn pa khng nguyn t cng</p><p>nhau vi p l cc s khng vt qu pa1 v chia ht cho p. C ngpa1 s nh vy. Do tn ti papa1 s nguyn nh hn pa v nguynt cng nhau vi pa. Vy, (pa) = pa pa1.V d 1.6. (125) = </p><p>(53)= 53 52 = 100 ; (210) = 210 29 = 525.</p><p>Tnh cht 5. Nu m,n l cc s nguyn dng nguyn t cng nhau</p><p>th (mn) = (m).(n).</p><p>Chng minh. Ta vit cc s nguyn dng khng vt qu mn thnh</p><p>bng sau:</p><p>1 m+ 1 2m+ 1 ... (n 1)m+ 12 m+ 2 2m+ 2 ... (n 1)m+ 23 m+ 3 2m+ 3 ... (n 1)m+ 3... ... ... ... ...</p><p>m 2m 3m ... mn</p><p>By gi gi s r l mt s nguyn khng vt qu m. Gi s (m, r) =</p><p>d &gt; 1. Khi , khng c s no trong dng th r nguyn t cng nhau</p><p>vi mn, v mi phn t ca dng u c dng km + r, trong </p><p>1 k n 1, d | (km+ r), v d | m, d | r.Vy, tm cc s trong bng m nguyn t cng nhau vi mn, ta</p><p>ch cn xem cc dng th r vi (m, r) = 1. Ta xt mt dng nh vy, n</p><p>cha cc s r,m + r, ..., (n 1)m + r. V (r,m) = 1 nn mi s nguyntrong dng u nguyn t cng nhau vi n. Nh vy, n s nguyn trong</p><p>dng lp thnh h thng d y mul n. Do c ng (n) s</p><p>trong hng nguyn t cng nhau vi n. Do cc s cng nguyn t</p><p>cng nhau vi m nn chng nguyn t cng nhau vi mn.</p><p>V c (m) dng, mi dng cha (n) s nguyn t cng nhau vi</p><p>mn nn ta suy ra (mn) = (m)(n).</p><p>Kt hp hai tnh cht trn, ta c tnh cht sau:</p><p>Tnh cht 6. Gi s n = pn11 pn22 ...p</p><p>nkk l phn tch n ra tha s nguyn</p><p>t. Khi :</p><p> (n) = n</p><p>(1 1</p><p>p1</p><p>)(1 1</p><p>p2</p><p>)...</p><p>(1 1</p><p>pk</p><p>).</p></li><li><p>9Chng minh. V l hm c tnh cht nhn nn nu n c phn tch</p><p>nh trn, ta c: (n) = (pa11 )(pa22 )...(p</p><p>akk ).</p><p>Mt khc: (pajj</p><p>)= p</p><p>ajj paj1j = pajj</p><p>(1 1pj</p><p>), j = 1, 2, ..., k.</p><p>Vy</p><p> (n) = pa11</p><p>(1 1</p><p>p1</p><p>)pa22</p><p>(1 1</p><p>p2</p><p>)...pakk</p><p>(1 1</p><p>pk</p><p>)= pa11 p</p><p>a22 ...p</p><p>akk</p><p>(1 1</p><p>p1</p><p>)(1 1</p><p>p2</p><p>)...</p><p>(1 1</p><p>pk</p><p>)= n</p><p>(1 1</p><p>p1</p><p>)(1 1</p><p>p2</p><p>)...</p><p>(1 1</p><p>pk</p><p>).</p><p>Tnh cht 7. Gi s n l mt s nguyn dng. Khi :d|p (d) = n.</p><p>Chng minh. Tng trn y c ly theo cc c s ca n. Ta phn</p><p>chia tp hp cc s t nhin t 1 n n thnh cc lp sau y. Lp Cdgm cc s nguyn m, 1 m n, m (m,n) = d. Nh vy m thuc Cdnu v ch nu d l c chung ca m,n v (m/d, n/d) = 1. Nh vy, s</p><p>phn t ca Cd l cc s nguyn dng khng vt qu n/d v nguyn</p><p>t cng nhau vi n/d ; tc l Cd gm (n/d) phn t. V mi s nguyn</p><p>m t 1 n n thuc mt v ch mt lp Cd no (d = (m,n) nn n</p><p>bng tng ca s cc thnh phn trong cc lp Cd, d l c s ca n.</p><p>Ta c n =d|n(nd</p><p>).</p><p>1.2. Hm tng cc c s dng ca n</p><p>1.2.1. nh ngha</p><p>nh ngha 1.1. Hm tng cc c dng ca s t nhin n c k</p><p>hiu l (n).</p><p>V d 1.7. (12) = 1 + 2 + 3 + 4 + 6 + 12 = 28.</p><p>Ch 1.2. Ta c th biu din hm (n) di dng: (n) =d|nd</p></li><li><p>10</p><p>1.2.2. Cc tnh cht</p><p>B 1.1. Gi s m,n l cc s nguyn dng nguyn t cng nhau.</p><p>Khi , nu d l c chung ca mn th tn ti cp duy nht cc c</p><p>dng d1 ca m v d2 ca n sao cho d = d1.d2. Ngc li, nu d1 v d2l cc c dng tng ng ca m v n th d = d1.d2 l c dng ca</p><p>mn.</p><p>Chng minh. Gi s m,n c phn tch ra tha s nguyn t nh sau:</p><p>m = pm11 pm22 ...p</p><p>mss ; n = q</p><p>n11 q</p><p>n22 ...q</p><p>ntt .</p><p>V (m,n) = 1 nn tp hp s nguyn t p1, p2, ..., ps v tp hp cc</p><p>s nguyn t q1, q2, ..., qt khng c phn t chung. Do phn tch ra</p><p>tha s ca mn c dng: mn = pm11 pm22 ...p</p><p>mss .q</p><p>n11 q</p><p>n22 ...q</p><p>ntt .</p><p>Nh vy, nu d l mt c chung ca mn th d = pe11 pe22 ...p</p><p>ess .q</p><p>f11 q</p><p>f22 ...q</p><p>ftt ,</p><p>trong 0 ei mi(i = 1, 2, ..., s) ; 0 fi ni(i = 1, 2, ..., s).t: d1 = p</p><p>e11 p</p><p>e22 ...p</p><p>ess , d2 = q</p><p>f11 q</p><p>f22 ...q</p><p>ftt . R rng d = d1d2 v (d1, d2) = 1.</p><p>Ngc li, gi s d1 v d2 l cc c dng tng ng ca m v n.</p><p>Khi :</p><p>d1 = pe11 p</p><p>e22 ...p</p><p>ess trong , 0 ei mi(i = 1, 2, ..., s)</p><p>d2 = qf11 q</p><p>f22 ...q</p><p>ftt trong , 0 fi mi(i = 1, 2, ..., t).</p><p>S nguyn d = d1d2 = pe11 p</p><p>e22 ...p</p><p>ess .q</p><p>f11 q</p><p>f22 ...q</p><p>ftt .</p><p>R rng l c ca mn = pm11 pm22 ...p</p><p>mss .q</p><p>n11 q</p><p>n22 ...q</p><p>ntt</p><p>v ly tha ca mi s nguyn t xut hin trong phn tch ra tha s</p><p>nguyn t ca d b hn hoc bng ly tha ca s nguyn t trong</p><p>phn tch ca mn.</p><p>B 1.2. Gi s p l s nguyn t, a l s nguyn dng. Khi :</p><p> (pa) =(1 + p+ p2 + ...+ pa</p><p>)=</p><p>pa+1</p><p>p 1 (pa) = a+ 1</p></li><li><p>11</p><p>Chng minh. Cc c ca pa l 1, p, p2, pa. Do , pa c ng a + 1 c</p><p>dng, (pa) = a+1. Mt khc, (pa) = 1+p+p2+ ...+pa =pa+1 1p 1 .</p><p>nh l 1.1. Gi s f l mt hm c tnh cht nhn. Khi hm</p><p>F (n) =d|nf(d) cng c tnh cht nhn.</p><p>Chng minh. Ta s ch ra rng num,n l cc s nguyn dng nguyn</p><p>t cng nhau th F (mn) = F (m).F (n). Gi s (m,n) = 1, ta c:</p><p>F (mn) =d|mn</p><p>f(d).</p><p>V (m,n) = 1 nn theo b 1.1, mi c s ca mn c th vit duy</p><p>nht di dng tch cc c d1 ca m v d2 ca n v d1, d2 nguyn t</p><p>cng nhau, ng thi mi cp c s d1 ca m v d2 ca n tng ng</p><p>vi c d1.d2 ca mn. Do ta c th vit: F (mn) =d1|md2|n</p><p>f(d1d2).</p><p>V f l hm c tnh cht nhn v (d1, d2) = 1 nn</p><p>F (mn) =d1|md2|n</p><p>f(d1)f(d2) =d1|m</p><p>f(d1).d2|n</p><p>f(d2) = F (m).F (n)</p><p>Tnh cht 1. Hm (n) l hm nhn tnh, tc l: Vi mi s t nhin</p><p>n1, n2 nguyn t cng nhau th (n1.n2) = (n1).(n2)</p><p>Chng minh. T nh l 1.1 suy ra hm s (n) c tnh cht nhn. V</p><p>th ta c th vit cng thc ca chng khi bit phn tch thnh tha s</p><p>nguyn t ca n.</p><p>Tnh cht 2. Nu p l s nguyn t th (p) = 1 + p</p><p>Chng minh. c suy ra t B 1.2.</p><p>Tnh cht 3. Gi s n l s nguyn dng v c khai trin chnh tc</p><p>n = p11 p22 ...p</p><p>kk th (n) =</p><p>p1+11 1p1 1 .</p><p>p2+12 1p2 1 ...</p><p>pk+1k 1pk 1</p><p>Chng minh. Do hm c tnh cht nhn nn ta c</p><p>(n) = (pa11 ) (pa22 ) ... (p</p><p>ass ).</p></li><li><p>12</p><p>1.3. Hm tng cc ch s ca s t nhin n</p><p>1.3.1. nh ngha</p><p>nh ngha 1.1. Gi s n l mt s t nhin. Ta nh ngha S(n) l</p><p>hm tng cc ch s ca n, khi biu din trong h thp phn.</p><p>1.3.2. Cc tnh cht</p><p>Vi n l s nguyn dng. Ta c:</p><p>Tnh cht 1. S(n) n (mod 9).Chng minh. Gi s trong biu din thp phn, s nguyn dng n</p><p>c dng: n = kk1...210 |10Khi y</p><p>n = 0 + 101 + 1022 + ...+ 10</p><p>k1k1 + 10kkS(n) = 0 + 1 + 2 + ...+ k1 + k</p><p>V th</p><p>n S(n) = 91 + 992 + ...+ 99...9 (k - 1) s 9</p><p>k1 + ...+ 99...9 k s 9</p><p>k. (1.1)</p><p>T (1.1) suy ra [n S(n)] ... 9 hay S(n) n (mod 9), suy ra iu phichng minh.</p><p>Tnh cht 2. 0 &lt; S(n) nTnh cht 3. S(n) = n 1 n 9Chng minh. Ta c n = kk1...210 |10 . V n &gt; 0 nn k &gt; 0.Ngoi ra i {0, 1, 2, ..., 9} vi mi i = 1, 2, ..., k.T , do S(n) = k + k1 + ...+ 1 + 0 suy ra S(n) &gt; 0.Li thy t (1.1) th S(n) n v S(n) = n 1 = 2 = ... = k =0 0 &gt; 0 0 {1, 2, ..., 9}. l iu phi chng minh.Tnh cht 4. S(m+ n) S(m) + S(n), vi mi m,n nguyn dng.Chng minh. Gi s trong h thp phn, n v m ln lt c dng:</p><p>n = kk1...10 |10m = kk1...10 |10</p></li><li><p>13</p><p>Khng gim tng qut, ta c th cho l n m k s. Ta c th vitli m di dng sau y m = 00...0 </p><p>(k - s) s 0</p><p>ss1...10 |10</p><p>t i=</p><p>{i vi i = 0,1,2,...,s</p><p>0 vi i = s + 1,...,k.</p><p>V th lun lun c th coi n v m c cng loi biu din sau:</p><p>n = kk1...210 |10m = kk1...210 |10</p><p>trong 0 i, i 9, vi mi i = 0, 1, 2, ..., k v i, i nguyn.T...</p></li></ul>