JPJC H3 Math Prelim 2024 Solutions

1 9820/01/PRELIM/2023 [Turn over JPJC J2 Preliminary Examination 2024 H3 Mathematics Solutions Q1 (a)i) Let ( )gcd ,a b d= | and |d a d b Since | and | ,d a d b ( ) 1 2 1 2 2 1 2 1 and , where and are integers | k d a k d b k k b a k d k d k k d d b a  = = − = − = − − Since | and |d a d b a − , ( )| gcd ,d a b a − Let ( )gcd ,a b a e−= | and |e a e b a− ( ) ( ) ( ) 3 4 3 4 3 4 3 4 34 and e , where and are integers e | k e a k b a k k a b a k e k k k e b k k e eb  = = − + − = + = + =+  Since | and e |e a b , ( )| gcd ,e a b | and | , then d e e d d e = ( ) ( )gcd , gcd ,a b a a b − = (ii) ( ) ( ) ( )gcd 72,120 gcd 72,120 72 gcd 72, 48= − = Similarly, ( ) ( )gcd 48,72 gcd 48, 24 24== (b) Let ( )( ) ( )( )12gcd ,gcd , and gcd gcd , ,a b c d a b c d== Since ( )( ) 1gcd ,gcd ,a b c d = , ( ) ( ) ( )( ) 11 1 1 1 11 1 12 | a and | gcd , | a and | and | | gcd , and | | gcd gcd , , | d d b c d d b d c d a b d c d a b c dd     

2 ( )( ) ( ) ( ) ( )( ) 2 22 2 2 2 22 2 21 Similiarly, since gcd gcd , , , | gcd , and | | and | and | | and | gcd , | gcd ,gcd , | a b c d d a b d c d a d b d c d a d b c d a b c dd =      Since 2 1 1 2 1 2| and | , d d d d d d =

3 9820/01/PRELIM/2023 [Turn over 2(i)(a) Equivalent to 1 3 4 5 2 13x x x x x+ + + + = ,  0ix + Number of ways 13 4 4 += 2380= (i)(b) Equivalent to 1 3 4 5 2 13x x x x x+ + + + = ,  1 1, 2,3x  , ix + Equivalent to 1 3 4 5 2 8y y y y y+ + + + = ,  0,1,2iy  ,  0iy + [so 1 coin of each type] Number of ways 84 4 += Complement is equivalent to 1 3 4 5 2 8z z z z z+ + + + = , 3iz  ,  0iz + [at least 3 5 cent coins] equivalent to 1 3 4 5 2 5w w w w w+ + + + = ,  0iw + Required number of ways 54 4 += Required number of ways 8 4 5 4 44 ++   =−       369= (ii)(a) Number of ways 125 4 83886080=  = (ii)(b) Number of ways 13 13 13 13 135 5 5 55 4 3 2 11 2 3 4        = −  +  −  +                901020120=

4 Q3 3(i) 1 1 () p p p p p i i i p xix yy xy − − = ++  =+  Note that ( 1) ( 1) ! p pipp i i  −  −= +   For 11 ip  − , since ip and p is prime, thus ( 1)!| ( 1)i p i p− −+ and p is a factor of p i   . Accordingly, ( ) (mod )p p p pxyxy ++ 3(ii) Let aP be the proposition that ( mod )pa a p for all positive integers a . Clearly, 11p = . Thus 1P is true. Suppose kP is true for some k + . Consider 1kP + . ( 1) (mod ) (mod ) (by inductio 1 n hypothesi1 s) pp p k p k k+ + + Thus 1kP + is true. Since 1P is true and kP is true  1kP + is true, by mathematical induction, ( mod )pa a p for all positive integers a . 3(iii) Since n is not a multiple of 4, we must have 4n k r=+ for some k + and 1,2,3r= . Using (ii), for ap , we must have 11 (mod ) 1 (