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 (mod )ppa p pa a a−−    Now for 1,2,3,4i= , 44 (mod 51 )n k r k r k r riii i i i += == 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 =  .

5 9820/01/PRELIM/2023 [Turn over 4(a) ( ) ( )( ) ( ) ( )( ) ( )( ) 2 a p b q c r a b c p q r a p b q c r a b c p q r ap bq cr a b c p q r  +  +   + + + +  +  +   + + + + + +  + + + + (b) 2 11 2 1 (1)22 x y xy xy xy x x x x y y xy + +  = −+ Similarly, 1 (2)2 yy y z z −+ 1 (3)2 zz z x x −+ (1) (2) (3) ( ) ( ) ( ) ( )( )( ) ( )( )( ) 111 222 1 8 1 8 x y z x y z x y y z x z y z x xyz x y z x y y z x z y z x xyz x y y z x z     + + +   + + + + + + (c)

6 ( )( ) ( )( )( ) ( )( ) ( )( )( ) ( )( ) ( )( )( ) ( )( ) ( )( ) ( )( ) ( )( )( ) ( ) ( ) ( )  ( ) ( ) ( )  ( )( )( ) ( )( ) ( )( )( ) ( )( ) ( )( )( ) 2 2 22 2 2 2 222 22 222 22 22 x y z z x y x z x y z x y y zx y z x y y z z x x y y z z x y z x y z x z x x y y z xy x y z z x y x z x y z x y y z x y z y x x z z x y z x x y y z z x x y y z z x xy yz xz x y z x y y z z x xy yz xz x y z x y y z z zx y + + + + + + + + + + + ++ + + + ++ + = + ++ + + + + + + + + + + = + + + + + + + + + += + + + ++ + + + + + ++= + + + + = ++  ( )( )( ) ( )( )( ) ( )( )( ) ( )( )( ) ( )( )( ) 22 2 2 1 2 2 1 322 22 3 1 8 y y z z x xyz x y y z z x x y y z z x xyz xy xyz x y y z y z x x zz + + + + + + + + + + += ++ + + = = ++  + + = 5(i) Eugene does not do a threshold run on two consecutive days and he does not do a recovery run for more than two consecutive days. Call this condition (*). For 1na + , Day 1 is a threshold run. Day 2 cannot be a threshold run. Days 2 to 1n+ is a sequence of n runs satisfying (*) where Day 2 is a tempo run or recovery run. By Addition Principle, 1nn na b c+ =+ For 1nb + , Day 1 is a tempo run. Days 2 to 1n+ is a sequence of n runs satisfying (*) where Day 2 can be any run. By Addition Principle, 1 nn n nb a b c+ = + + For 2nc + , Day 1 is a recovery run.

7 9820/01/PRELIM/2023 [Turn over Case 1: Day 2 is a threshold run Days 2 to 2n+ is a sequence of 1n+ runs satisfying (*) where Day 2 is a threshold run. Case 2: Day 2 is a tempo run Days 2 to 2n+ is a sequence of 1n+ runs satisfying (*) where Day 2 is a tempo run. Case 3: Day 2 is a recovery run Day 3 cannot be a recovery run. Case 3A: Day 3 is a threshold run Days 3 to 2n+ is a sequence of n runs satisfying (*) where Day 3 is a threshold run. Case 3B: Day 3 is a tempo run Days 3 to 2n+ is a sequence of n runs satisfying (*) where Day 3 is a tempo run. By Addition Principle, ( )2 1 1n n n n nc a b a b+ + += + + + (ii) Doing a replacement yields 3 4 3 3 32 4 21 3 1 3 (1) (2) (3) nn n n n n n n n n n n a b c b a b c c a b a b + + + + + + + + + ++ + =+ = + + = + + + Sub (1) into (2), 4 3 4 (4)n n nb a a+ + +=+ Doing a replacement yields 11 2 1 2 3 2 3 (5) (6) (7) n n n n n n n n n b a a b a a b a a ++ + + + + + + =+ =+ =+ Sub (5) and (6) into (3), ( ) ( )3 2 1 2 1 1 212 3 (8) n n n n n n n n n n c a a a a a a a a a + + + + + + ++ = + + + + + = + + Sub (7) and (8) back into (1), ( ) ( )2 3 2 1 4 23n n n n n na a a a a a++ + ++= + + + + 3 2 133n n n na a a a+ + += + + +

8 (iii) Method 1: Recurrence 1 2 3 1 1 2 2 1 2 3 6 a a a = =  = =   = 4 1st day THR 1st day 5 THR 2nd day TEM 2nd day REC 2 3 3 2 3 2 15 15 3(6 2) 1 40 (shown) a a = + + + + + = = + + + = Method 2: For 5a , Day 1 is a threshold run Day 2 can be only be a tempo run or recovery run. Case 1: Day 2 is tempo run Case 1A: Day 3 is threshold run, Day 4 is tempo or recovery run, Day 5 is any run No. of ways 2 3 6=  = Case 1B: Day 3 is tempo run, Day 4 is threshold run, Day 5 is tempo or recovery run No. of ways 2= Case 1C: Day 3 is tempo run, Day 4 is tempo or recovery run, Day 5 is any run No. of ways 2 3 6=  = Case 1D: Day 3 is recovery run, Day 4 is threshold run, Day 5 is tempo or recovery run No. of ways 2= Case 1E: Day 3 is recovery run, Day 4 is tempo run, Day 5 is any run No. of ways 3= Case 1F: Day 3 is recovery run, Day 4 is recovery run, Day 5 is threshold or tempo run No. of ways 2= Case 2: Day 2 is recovery run Case 2A: Day 3 is threshold run, Day 4 is tempo or recovery run, Day 5 is any run No. of ways 2 3 6=  = Case 2B: Day 3 is tempo run, Day 4 is threshold run, Day 5 is tempo or recovery run No. of ways 2= Case 2C: Day 3 is tempo run, Day 4 is tempo or recovery run, Day 5 is any run

9 9820/01/PRELIM/2023 [Turn over No. of ways 2 3 6=  = Case 2D: Day 3 is recovery run, Day 4 is threshold run, Day 5 is tempo or recovery run No. of ways 2= Case 2E: Day 3 is recovery run, Day 4 is tempo run, Day 5 is any run No. of ways 3= By Addition Principle, no. of ways 40= Method 3: For 5a , Day 1 is a threshold run Day 2 can be only be a tempo ru