[ 47 ] then T ( m ) is true for all n a. 2. The greatest integer function : [ ] is defined by selling [ x]

Transcription

1 [ 47 ] Number System 1. Itroductio Pricile : Let { T ( ) : N} be a set of statemets, oe for each atural umber. If (i), T ( a ) is true for some a N ad (ii) T ( k ) is true imlies T ( k 1) is true for all k a, the T ( m ) is true for all a.. The greatest iteger fuctio : [ ] is defied by sellig [ x] the greatest iteger ot exceedig x, for every real x.. If a/ b ad a/ c, the a/ b qc (liearity roerty). 4. Euclid's Algorithm : The a ad b be two o-zero itegers. The ( a, b ) [gcd of ( a, b )] exists ad is uique. Also, there exists itegers m ad method ( a, b) am b. 5. Cogruecies : Let a, b i be itegers, m 0. The we say that a is cogruet to b modulo m if, m /( a b). We deote this by a b(mod m). 6. Let a b(mod m) ad c d (mod m) The (i) a c b d(mod m) (ii) a c b d(mod m) (iii) ac bd(mod m) (iv) a qc b qd (mod m) for all itegers ad q. (v) a m b mod m for all ositive iteger m. (vi) f ( a) f ( b)(mod m) for every olyomial with iteger coefficiets. 7. A iteger x 0 satisfyig the liear cogruece ax b(mod m) has a solutio. Further more, if x 0 is a solutio, the the set of all solutios is recisely ( x0 km) : ( Z). 8. Let N be a ositive iteger greater tha 1, say N a b c... where a, b, c... are distict (differet rimes ad, q, r... are ositive itegers. The umber of ways i which N ca be resolved ito two factors is q r

2 [ 48 ] 1 ( 1)( 1)( 1)... q r 9. Number of ways i which a comosite umber ca be resolved ito two factors, which are rime to 1 each other, is where is the umber of distict rime factors i the exressio for N. 10. Let N be a ositive iteger quarter tha 1 ad let N a b c... where a, b, c... are distict rimes ad, q, r... ositive itegers. the the sum of all the divisors i the roduct is equal to 1 q1 r1 q a 1 b 1 c 1.. a 1 b 1 c the highest ower of rime which is cotaied i! is equal to where [ ] is the greatest iteger fuctio Euler's Totiet Fuctio : Let N be ay ositive iteger 1. the umber of all ositve itegers less tha N ad rime to it is deoted by ( N). It is obvious () 1, (), (4), (5) 4, (6)... The fuctio is called Euler's Totiet fuctio. If a, b,... are rime to each other, the ( ab) ( a). ( b) or ( abcd...) ( a) ( b) ( c)... If N a b c... where a, b, c are distict rimes ad, q, r are ositive itegers, the q r q r ( N) N a b c Euler's Theorem : If x be ay ositive iteger rime to N. The x ( N ) 1(mod N) 1. Fermat's Little Theorem : If is a rime ad is rime to the 14. Wilso's Theorem : If is a rime, their Examle : ( 1)! 0(mod ) Coversely, if ( 1)! 1 0 ( mod ), their is a rime. Calculate 09 5 (mod 41). Sice 41 is rime ad (5, 41) 1, therefore by Fermat s (mod 41) by divisio algorithm, 09 (50 40) (mod ) (5 ) (mod 41) 9 5 (mod 41)

3 [ 49 ] To calculate 9 5 (mod 41), we first calculate 5 (mod 41) where, 4, 8, 16, (mod 41) (mod 41) (mod 41) (mod 41) 5 16(mod 41) ( 16) (mod 41). (mod 41) Questio 1. Fid the largest ositive iteger such that Usig modulo ( 10) umbers, we see that 10 0[mod ( 10)] i.e., 10[mod( 10)] ( 10) [mod( 10)] 1000([mod( 10)] 100 is divisible by ( 10). 100 ( )[mod( 10)] 900[mod( 10)] Now, we wat ( 10) to divide 100, imlyig that ( 10) should divide 900. The largest such is , as ( 10) caot be greater tha ad the greatest divisor of 900 is 900. so the largest ositive iteger, such that 100 is divisible by ( 10) is 890. Note : has 7 divisors a each divisor greater tha 10, gives a corresodig value for they are, 5, 8, 10, 15, 0, 6, 5, 40, 50, 65, 80, 90, 140, 170, 15, 90, 440 ad 890. Questio. Whe the umbers ad are divided by a certai three digit umber, they leave the same remaider. Fid this largest such divisor ad the remaider. How may such divisors are there? Let the divisor be d a the remaider be r. The by Euclidea Algorithm, we fid dq1 r (1) ad dq r ()

4 [ 50 ] by subtractig Eq. () from Eq. (1), we get d is a three digit divisor of d( q q ) 1 therefore, ossible values of d are 891, 594, 97, 198. Hece largest three digit divisor is 891 ad the remaider is 177. Questio. Fid the umber of all ratioal umbers m rime ad (iii) m. 5!. It is give that m such hat (i) 0 1, (ii) m ad are relatively m 5! Thus 5! is the roduct of owers of 9 rime umbers. The umber of ways i which 5! ca be writte as the roduct of two relatively rime umbers m ad is less tha 1. There are 9, which leads to 8 such fractios. 9 factors, exactly half of which, are such that m is Questio 4. Determie all ositive itegers for which 1 is divisible by. 1 1 If is odd, the ( 1) is a factor. Thus for all odd values of, 1 is divisible by. Aliter : 1 (mod ) 1(mod ) (mod ) ad so o m 1 (mod ) ad m m m ( ) 4 1 mod () mod if is odd. ad mod if is eve. 1 is divisibly by if is a odd umber Questio 5. Prove that [ x] [ x] [4 x] [8 x] [16 x] [ x] 145 has o solutio. 145 x x 4x 8x 16x x 6x x Whe x 196, the L.H.S. of the give equatio becomes x

5 Questio 6. [ 51 ] 1 Cosider x i the iterval 195, 196. the L.H.S. exressio of the give equatio Whe x 195, the L.H.S. is less tha 14. for o value of x, The give equality will be satisfied. Three cosecutive ositive itegers raised to the first, secod ad third owers resectively, whe added, make a erfect square, the square root of which is equal to the sum of the three cosecutive itegers. Fid these itegers. Let ( 1),, 1 be the three cosecutive itegers. The 1 ( 1) ( 1) ( ) 9 Questio 7. Show that ( 1)( 4) 0 0 or 1 or 4, but 0 ad 1 will make the cosecutive itegers 1, 0, 1 ad 0, 1, ad, which cotradicts the hyothesis that the cosecutive itegers are all greater tha zero. Hece 4, corresodig to which the cosecutive itegers are, 4 ad is divisible by We shall make grous of the terms of the exressio as follows : ( ) ( 1995 )... ( ) Here each bracket is of the form But ( a b ) 1997 for all i. i i 1 1 i bi ( a ) is divisible by ( a b ). Each bracket ad hece, their sum is divisible by Questio 8. A four digit umber has the followig roerties : (a) (b) (c) It is a erfect square The first two digits are equal The last two digits are equal. Fid all such umbers. Let N aabb be the reresetatio of such a umber. 1 a 9, 0 b 9 The N 1000a 100a 10b b 1100a 11b 11(100 a b) i i

6 [ 5 ] Questio 9. Sice N is a erfect square, ad 11 is a factor of N aob 100a b must be a multile of 11, i.e., aob should be divisible by 11 where a, o ad b are digits of 100a b ad hece b 11 a. The last digits of a erfect square where both the digits are equal is oly 44. So b 4 a 7 This is the oly solutio. Show that N 7744 is the oly ossibility is divisible by ( 1) (mod 11) So ( ) ( 1) (mod 11) 1 (mod 11) (mod 11) It is a multile of 11. Questio 10. The equatio x x 9 0 has ratioal roots, where ad q are itegers. Prove that the roots are itegers. x 4 q, sice the roots are ratioal, If is eve, 4q is a erfect square. ad 4q are eve ad hece 4q is a eve iteger ad hece, 4q, is a eve iteger ad hece, 4q is a iteger. If is odd, 4q ( 4 q) is odd ad is a iteger ad hece, the result. 4q is a eve iteger ad hece, Questio 11. Fid all airs of atural umbers, the differece of whose squares is 45. Let x ad y be the atural umbers such that ( x y)( x y) 45 x y 45 where x y. so, both ( x y) ad ( x y) are the divisors of 45 ad x y x y, where x ad y are ositive itegers. So, x y 1, x y 45 (1)

7 Solvig (1), () ad (), we get (a) x, y (b) x 9, y 6 (c) x 7, y [ 5 ] x y, x y 15 () x y 5, x y 9 () so, the airs of umbers satisfyig the coditio are (, ), (9, 6), (7, ). Questio 1. Show that ay circle with cetre (, ) caot ass through more tha oe lattice oit. [Lattice oits are oits i cartesia lae, whose abscissa ad ordiate both are itegers.] If ossible, let ( a, b), ( c, d ) be two lattice oits o the circle with (, ) as cetre ad radius r. ( a ) ( b ) r ( c ) ( d ) a b c d ( a b) ( c d) ( a c) ( b d) Sice ad are irratioal umbers ad, a, b, c ad d are itegers ( a c) ( b d) is irratioal as ad are ulike irratioal umbers ad hece, addig k ad l where k ad l are itegers does ot give a ratioal umber. But the left had side a b c d is a iteger. It is a cotradictio ad thus, the circle with (, ) as cetre ca ass through utmost oe lattice oit. [Note : You may kow that the equatio of the circle whose cetre is ( g, f ) ad radius r give by for the circle If it asses through the origi, the x y gx fy g f r 0 x y x y (5 r ) 0. 5 r 0 or r 5, so the oe lattice oit that lies o the circle with cetre (, ) ad radius 5 is (0, 0). But it is ot ecessary that there exists at least oe lattice oit for circles with such cetres.] Questio 1. Fid all ositive itegers for which 96 is erfect square. Let 96 k, where k N. The Clearly k k ( k )( k ) 96 ad hece, k k 0.

8 [ 54 ] Sice is the oly odd factor, both k ad are itegers. We must have k ad k both to be either eve or odd. (If oe is odd ad the other eve, the k ad do ot have iteger solutios). Also both k ad k caot be odd as the roduct is give to be eve. So the differet ossibilities for k, k are as follows. k k 48 (1) k 4 k 4 () k 6 k 16 () k 8 k 1 (4) So, solvig searately Eqs. (1), (), () ad (4), we get, 10, 5,. So, there are exactly four values for which gives 10 gives 5 gives gives 96 is a erfect square Questio 14. There are ecklaces such that the first ecklace cotais 5 beads, the secod cotais 7 beads ad i geeral, the tth ecklace cotais i beads more tha the umber of beads i ( i 1) th ecklace. Fid the total umber of beads i all the ecklaces. Let us write the sequece of the umber of beads i the 1st, d, rd,..., th ecklaces. S S 5, 7, 10, 14, 19,... ( 1) (4 1), (4 ), (4 6), (4 10), (4 15),..., 4 Total umber of beads i the ecklaces times ( 1) Sum of the first triagular umbers. 1 4 ( ) 1 4 ( ) 1 ( 1) ( 10) 1 ( 1) 4 6 ( 1) ( 1) ( 1) [48 ( 1)( )] 1

9 [ 55 ] [ 6]. 6 Questio 15. If a, b, c are ay three itegers, the show that by 7. Le us fid the value of As a (mod 7) is 0, 1,,, 4, 5 or 6, a (mod 7) for ay a Z. abc( a b )( b c )( c a ) is divisible a (mod 7) will be oly amog 0, 1 or 6. Now, if 7 divides oe of a, b, c, the give exressio is divisible by 7. If ot, the (mod 7) will be oly amog 1 ad 6. Hece, two of them must be the same, say b mod 7). Questio 16. Show that ( a b ) (mod 7) = 0. The give exressio is divisible by is divisible by 5. The give exressio may also be writte as (1 4 ) ( ) 5 a, b, c a ad Now (1 4)( ) 5 (say) Similarly Exressio is equal to where, q, r are itegers q say 5 5q 5 5 r (ay) 5 divides the give eumessio as 5r is divisible by 5. Questio 17. Ram has four differet cois with values as show o the right. Suose you had just oe of each of these cois (4). How may differet amouts (value) ca be made usig oe or more of the four differet cois? Exlai. Let us list the ossible amouts : Coi : This roduces 1,, 4, 8. Coi : They roduce = (1 ); 5 = (1 4); 9 = (1 8) 6 = ( 4); 10 = ( 8); 1 = (4 8). Coi : They roduce 7 = (1 4), 11 = (1 8), 1 = (1 4 8); 14 = ( 4 8);

10 Questio 18. Show that Let [ 56 ] 4. Coi ; They roduce 15 = (1 4 8) There are 15 differet amouts that ca be made. These amouts are the first 15 coutig umbers is a ositive iteger divisible by (say) 9 99 Now 16 () (81) sice 19 ad 1 are both odd oe with add, their differece is eve ad hece divisible by we have to show that divisible by 81. Now (18 1) (9)18 1 (mod 81) 1675(mod 81) 55(mod 81) Also (1 1) (99)1 1 (mod 81) 1189(mod 81) 55(mod 81) Questio 19. Show that (mod 81) Thus 16 ()(81) divides ] ad without usig tables/calculatio ; Questio 0. Let be a rime umber >. What is the remaider whe ad rime so is odd. ( 1) ad ( 1) are both eve i.e., / 1 ad / ; is divided by 1? 4/ 1 (1) Also, ( 1),, ( 1) are cosecutive itegers. Oe of them must be divisible by. But / / 1 () 1/ 1 from (1) ad () whe is divided by 1, the remaider is 1.