ASR 2025 Prelim (Answers)

2025 ASRJC H23 Math Preliminary Exam Solutions 1 Solution: Consider the vectors a b c                  and x y z                  . cos a x a x b y b y θ c z c z                   where θ is the angle between vectors a b c                  and x y z                  .   2 2 2 2 2 2 cosax by cz a b c x y z θ       2 2 2 2 2 2 2 2 cosax by cz a b c x y z θ  Since 2cos 1θ  , Therefore      2 2 2 2 2 2 2ax by cz a b c x y z  [M1] [B1] For equality to hold, 2cos 1 cos 1 0 or .θ θ θ π  Therefore a necessary and sufficient condition is a b c                  is parallel to x y z                  . [A1] (i) ( ) ( )( ) ( ) ( ) 22 2 2 4 4 4 2 2 2 22 2 2 2 2 2 2 2 2 2 (1) (1) (1) 1 1 1 3 ( ) 2 ( ) ( ) ( ) a b c a b c a b c a b c ab bc ca + +  + + + + + +  + + − + + ( ) ( ) ( ) 22 2 2 2 2 2 22 2 2 2 2 2 2 6 ( ) ( ) ( ) ( ) ( ) ( )3 a b c ab bc ca abc ab bc ca  + +  + + ++   + + [B1] [M1] [M1] (ii) Let , 2 , and 8, 4, 1.x p y q z r a b c  LHS =   2 28 8 243 59049p q r  RHS =       2 2 2 2 2 28 4 1 4 81 729 59049p q r  Therefore these assigned values and variables satisfy      2 2 2 2 2 2 2ax by cz a b c x y z  Therefore 2 p q r                  is parallel to 8 4 1                  [B1] [M1] [B1]

ie 8 24 1 p qk r                   ie 8 ,2 4 ,p k q k r k  Substitute into 8 8 243p q r  , 64 16 243 81 243 243 381 k k k k k    24, 6, 3p q r  [M1] [A1]

2

3 Solution: (i) For 2P , the train has 2 seats.     22 1 sits in his assigned seat sits in his assigned seat 1 2 P P T PT          33 1 1 2 2 1 sits in his assigned seat sits in his assigned seat sits in S sit s in S 1 1 1 3 3 2 1 2 P P T P T P T T             [A1] [A1] (ii) If passenger 1T sits in kS , then 2 3 1, ,...., kT T T  will sit in their assigned seats as they are all available for them. For kT , since his assigned seat kS has been taken, he would have to choose randomly from the remaining  11n k n k  seats. The situation is the same as a train having 1nk  seats and the first passenger choosing his seat randomly. So we have  11 sits in | sits in n n k n kP T S T S P               1 1 11 1 1 1 1 1 1 1 2 sits in sits in sits in | sits in sits in 1 sits in | sits in sits in | sits in sits in | sits in 1 sits in n n n n k k n n n k k k n n n k k n n n n n k k n P P T S T S P T S T S P T S P T S T Sn P T S T S P T S T S n PT                 1 1 1 2 | sits in 1 10 nn n nk k S T S Pn                  Let 1r n k  , then when 2, 1k r n  and when 1, 2.k n r  1 1 2 1 2 1 1 1 1 n n n k k n r r PP n Pn                    [A2] [B1] [B1] [M1] [AG1 ]

(iii)   3 4 2 3 2 1 1 1 1 1 11 1 14 4 4 2 2 2 r r P P P P             Conjecture: 1 2 nP  We have already proven 1 1 2P  ie the result is true for 1.n  Assuming the result is true for 2,....., .nk  ie 12 1... 2 kP P P  . [A1] [B1] [B1]   1 2 2 1 11 11 112 11 1112 11 12 1 2 k kr r k r PP k k kk k k                                          Hence by Mathematical Induction, 1 2 nP  is true for all positive integer n. [M1] [A1] 4 Solution (a)Since    1 1 1 1n mp m p p              1 1 ( 1) 1 1 1 1 1 11 1 mod 1(mod ) 1(mod ) 1(mod ) since 1(mod ) by Fermat's Little Theorem 1(mod ) 1 mod since 1(mod ) (mod ) n m p p mp p mp p mp m p p xp xp x x p x p x p xp x p x p x x p               [B1] [B1] [B1] (b) 12 ..... kn p p p     If   1 1 modnxn   then 1 1nx sn   for some integer n

  1 121 ...n nx s p p p     1 1 modn ixp   for every ip   1 1 mod i i n pp ixp              From (a), we have   1 1 mod i n p ixp                   1 1 mod i n p ixp              for every ip From part (a),   1 1 mod i i n pp ixp                1 1 modn ixp   for every ip By Chinese Remainder Theorem,   1 1 modnxn   [B1] [B1] [B1] [B1] (c) Take any integer x with gcd( , ) 1.xn  Then for each |ipn , Fermat’s Little theorem gives   1 1 mod .ip ixp   Since    1 | 1 ,pn  we have  11n k p  for some integer k. Then       11 1 mod 1 mod .i kpnk iix x p p    This holds for every prime divisor ip of n. By Chinese Remainder Theorem,   1 1 modnxn   . [B1] [B1] [B1] 5 Solution: (i) There are n ways to pick a square from the 1st row. Suppose the square is in the jth column. “Block off” the squares in the 1st row and the squares in the jth column. The remaining (n – 1) squares can be picked from the ( n – 1) unblocked rows and the ( n – 1) unblocked columns in Sn-1 ways. Hence Sn = n Sn−1. [A1] (ii) If the square in row i and column j is selected, then map the integer i to the integer j. Conversely, if the integer i is mapped to the integer j, then select the square in row i and column j. Sn = n! [B1] [A1]

(iii) This is equivalent to a permutation of the integers {1, 2, 3, …, n} except that no integer is mapped to itself, e.g. the integer 1 cannot be mapped to itself (i.e. the 1 st slot) but to the other ( n – 1) slots. There are (n – 1) slots to choose from. Suppose 1 is mapped to the rth slot. There are (n - 1) position that 1 can occupy. Case 1 The integer r is mapped to 1st slot (i.e. 1 and r swop places) We just need to do a derangement of all the integers other than 1 and r, which can be done in Dn−2 ways. Case 2 The integer r is not mapped to 1 st slot (i.e. 1 and r don’t swop places). The rth slot is now occupied by “1”. For the time being, put “r” in the 1st slot. Now do a derangement of all the slots other than the rth slot. Since the “r” in the 1st slot cannot occupy the 1st slot as every integer must be mapped to another position, t his ensures that “r” is forced out of the 1 st slot. There are Dn−1 ways to do this derangement. Combining cases 1 and 2, we have Dn = (n – 1)(Dn−1+ Dn−2). D1 = 0 and D2 = 1. [B1] [B1] [B1] [A1] (iv) Let S be the set of all permutations. !Sn= Let kB be the set containing all permutations of n objects where the kth object is in its original position (kth position). Since ( )1!jBn=− for 1 jn ,we can choose any of the n integers to be fixed, ( ) 1 1!1 n j j nBn = =− . We can do the same for for jk and 1, j k n and similarly when three, four etc. objects are fixed. ( ) 12 1 1 12 ... ... 1 ... nn n j j k j k l j j k j k l n n D S B B B B B B B B B S B B B =    + = −     −  +  =−  + + −       [M1] [M1] [M1]

( ) ( ) ( ) ( ) ( ) ! 1 ! 2 ! ... 1 !12 1 1 1 1! 1 ... 11! 2! 3! ! n n n