CJC JPJC SAJC H3 Math 2024 Prelim Solutions

1 9820/01/PRELIM/2024 [Turn over Q1 Numbers and Proofs (a)(i) Let ( )gcd ,a b x= ; ( )gcd ,a b a y−= Since | and | ,x a x b | ( )x b a− as  and | | | ( )x a x b a x y −  ( )( )Similiarly, since | and | ( ), |y a y b a y a b a b− + − = as  a | |nd |y a y b y x Hence xy= , i.e. ( ) ( )gcd , gcd ,a b a b a=− (a)(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 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 =

Q2 Counting (a)(i) Equivalent to 1 3 4 5 2 13x x x x x+ + + + = ,  0ix + Number of ways 13 4 4 += 2380= (a)(ii) 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 + 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 += Therefore, required number of ways 8 4 5 4 44 ++   =−       369= (b)(i) Number of ways 125 4 83886080=  = (b)(ii) Number of ways 13 13 13 13 135 5 5 55 4 3 2 11 2 3 4        = −  +  −  +                901020120=

3 9820/01/PRELIM/2024 [Turn over Q3 Numbers and Proofs 3(i) 1 1 () p p p p i p i i p yix yy xx − − = ++  =+  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 1p− , we must have ( 1)n k p r= − + for some k + and 1,2,3r= . Using (ii), for ap , we must have 11 (mod ) 1 (mod )ppa p pa a a−−    Now for 1,2,3,4i= , ( 1) ( 1) (mod1 )n k p r k p r k r riii i i i p − + −== = Thus 44 11 (mod 5)nr ii ii ==  For 1,2,3,r= , 4 1 10,30,100r i i = = respectively. Thus 4 1 (mod 5)0n i i = 

Q4 Inequalities (a) ( ) ( )( ) ( ) ( )(