HCI 2026 H3 Math Prelim (Ans)

2026 H3 Math Prelim Mark Scheme Solution 1(a) nmmn , then nk mkmn knnk , then mk mnnk Therefore nk mnmk Then kmmk , shown 1(b) Let Pn be the statement ( ) 1d ed n n n axny a x nax −=+ , n +  For n =1, LHS = d eed ax axy axx =+ RHS = ( )1e axax+ =LHS  1P is true Assume Pk is true, ie. ( ) 1d ed k k k axky a x kax −=+ To prove 1Pk+ is true, ie. 1 1 1 d ( 1) ed k k k axk y a x k ax+ ++ = + + LHS = ( ) ( ) ( ) 1 1 1 1 1 dd edd ee 1e k k k ax k k k ax k ax k k ax y a x kaxx a x ka a a x k a + − + + + =+  = + + = + + =RHS Pk is true  1Pk+ is true, By principle of MI, Pn is true for all n +  2(a) [Method 1] Since !| !rn for ,0r r n +    , 1 1 1 1! 2! 3! ( 1)! !n nn + + + + −  is a positive integer for all positive integers n.

Solution [Method 2] 2 3 1 0 1 1 1 1! P P P P2! 3! ( 1)! ! n n n nnnn nn −− + + + + = + + + + −  Since Pn r +  for all ,0r r n +    , 1 1 1 1! 2! 3! ( 1)! !n nn + + + + −  is a positive integer for all positive integers n. (b) Since n +  , 0na  . ( )( ) ( )( )( ) ( )( ) ( )( )( ) 1 1 1 1 1 1 1 1 1 1 1 1 2 1 2 3 1 1 1 1 1 1 1 1 1 1 1 n n n n n n a n n n n n n n n n n n n n + + + + = + + ++ + + + + +  + + ++ + + + + + = − = =   Therefore, we have 10 na n . (shown) (c) From MF27, 1 1 1 1e 1 1 2! 3! ! ( 1)! nn= + + + + + + + +  ( )( ) ( )( )( ) 1 1 1 1!e ! 1 1 2! 3! ! ( 1)! 1 1 1! 2! 3! ! 1 1 1 1 1 2 1 2 3 2 1 1 12( !) ! 2! 3! ! n nn nn n n n n n n n n ann n = + + + + + + + + = + + + + + + + + + + + + + =+   +   + + + +    

Solution Since 2( !)n +  , and from (b) and (c), 01 na . 1 1 1Hence !e ! 2 2! 3! ! 1 1 1!2 2! 3! ! nn n a n n n  = + + + + +   = + + + +   !e !ena n n = −  (d) Assume that e is a rational number. So e pq= , ,pq +  . Consider nq= . !e !e !! ( 1)! ( 1)! ( 1)! ( 1)! 0 qa q q ppqq qq p q p q p q p q =−     =−        = − − −  = − − − = which is a contradiction since 0qa  . Hence e is irrational. Solution 3(a) ( ) ( ) ( ) 1 2 0 2 112 ( , ) 1 d 1 ' ' ( 2 ) 1 1 nm n m mn I m n x x x u x v x xu n x x v m +− =− = − = = − − = + 

( ) ( ) ( ) ( ) ( ) ( )( )( ) ( ) ( ) ( ) 1 2 0 111 1 122 0 0 1 122 0 1 122 0 1 122 0 1 122 0 ( , ) 1 d 1 ( 2 ) 1 d 11 2 1 d1 2 1 d1 2 1 1 1 d1 2 1 1 d1 2 ( , 1) ( , )1 nm mmnn nm nm nm nnmm I m n x x x xxx n x x xmm n x x xm n x x x xm n x x x xm n x x x x xm n I m n I m nm ++ − −+ − − − =− = − − − − ++ =− + =− + = − − −+ = − − −+ = − −+       221 ( , ) ( , 1)11 2 1 2 ( , ) ( , 1)11 2( , ) ( , 1) (shown)21 nn I m n I m nmm m n n I m n I m nmm nI m n I m nmn + = − ++ ++ =− ++ =− ++ 3(b) ( ) 2(3, ) (3, 1) (3, 1)3 2 1 2 1(3,1) (3,0)3 2 2 1(3, 2) (3,1) (3,0)4 4 3 3 3 2 1(3,3) (3, 2) (3,0)5 5 4 3 !2(3, ) (3,0) (3,0) ( 2)( 1)2! 2 nnI n I n I n nn II I I I I I I nI n I I nnn = − = −+ + + =     ==              ==            = = +++   1 3 0 1(3,0) d 4I x x == 2 1 1(3, ) (shown)( 2)( 1) 4 2( 2)( 1)In n n n n  = = + + + + 

3(c) ( )   ( ) ( ) ( ) ( ) ( ) 1 2 0 2 122 11 122 00 (0, ) 1 d 0,1 ,0 1 1 11 1 d 1 d (0, 1) (0, ) n nn nn I n x x xx xx x x x x I n I n + + =−   −   −  −  −  −  +    Solutions 4 (a)(i) Let iA be the set of distributions where student i receive the book they submitted. Total number of permutations !n= ( )1!iAn=− (Fix one student and permutate the rest) ( ) 1 2 3 ... ! ki i i iA A A A n k    = − By principle of inclusion and exclusion, ( ) ( ) ( ) 1 2 3 1 1 1 1 .. 1! !1 ! n n k k n k k A A A A n nkk n k − = − =     = − −  =−   ( ) ( ) ( ) 1 2 3 1 1 1 1 0 ! .. !!1 ! 1! 1 1 ! 1!1 ! nn n k k n k k n k k D n A A A A nn k n k n k − = − = = = −     = − − = + − =−     (a)(ii) Consider student 1, since he did not receive his book, there are 1n− choices for his book. We call this student k ’s book. Consider the book student k receives, Case 1: student k receives student 1’s book. Then we just arrange the other 2n− students: 2nD −

Case 2: student k does not receive student 1’s book. Now if we consider the other 1n− students other than student 1, this is a similar derangement where student k ’s book is removed and replaced by student 1’s book. Number of ways= 1nD − Hence, ( )( )121n n nD n D D −−= − + 4(b) (i) Group the integers into        1,36 , 2,35 , 3,34 ,.... 18,19 Since there are 18 such sets and 19 integers in X , at least 2 of the integers in X will be in the same set and that gives us the required a and b . (b)(ii) Consider the elements integers in S modulo 7 ,               1 2 3 4 5 6 0 1,8,15, 22, 29,36 2,9,16, 23,30 3,10,17, 24,31 4,11,18, 25,32 5,12,19, 26,33 6,13, 20, 27,34 7,14, 21, 28,35 S S S S S S S = = = = = = = We need to avoid an integer from 1S and 6S , 2S and 5S , 3S and 4S or 2 from 0S . Hence we can choose 6 5 5 1 17+ + + = integers at most (b) (iii) Set Y has 102 1 1023−= possible non-empty subsets. The largest possible sum is 27 28 ... 36 315+ + + = . The minimum is 1 . Hence, we have 1023 pigeons fitting into 315 pigeonholes. Thus, there are at least 2 subsets with the same sum. /{ }& /{ }X X Y Y X Y , two distinct subsets. 5(a) For k > 0 221 22k x k kx+  = 224 2 4 4k y k ky+   = 229 2 9 6k z k kz +   =

21 4 9 ( ) 12k x y z kx y z+ + + + +  21 4 9 12kkx y z+ + +  For all the equal sign to be held, 211 k x xxk=  = Similarly, 23,yz kk== Then 1 2 3 16 kkkk+ + =  = Therefore 21 4 9 12 6 6 36x y z+ +   − = . Where 1 1 1,,632x y z= = = 5(b) Let 11t x y xy= + + + 2 1 1 1 1 2 1( ) 2 1112 t x y x y x y x y x y y xx y xy xy xy x y yxxyxy xy x y = + + + + + +   += + + + + + + = + + + + + Since 1xy+= 2x y xy+ 10 4xy Consider xy as one variable, and the graph 1m xy xy=+ , if 10 4xy , The range of m will be 1 4,4 + 

Therefore 1 17 4xy xy+ And 2yxxy+ Therefore 2 171 4 2 2 10 4 10 t t  + + + =  Where x = y = ½. 5(c) (i) (ii) From Cauchy inequality, ( ) ( ) 2 2 2 2 2 0 1 2 0 1 2 2 2 ... ( 1) ... ( 1) 1 nnx x x x n x x x x a n b ab n + + + +  + + + + + +  + From Cauchy Inequality, ( ) ( ) 2 2 2 2 1 2 1 2 ... ... nnx x x n x x x+ + +  + + + ( ) ( ) 2 2 00a x n b x−  − …………………..(1) Then 22 00( 1) 2 0n x a x a nb+ − + −  Consider the discriminant  ( ) 22 22 2 4 4( 1)( ) 4 4 4 4 a n a nb na n b nb n nb b a = − + − =− + + = + − Consider equation 22 00( 1) 2 0n x a x a nb+ − + − = The roots are 2 0 2 0 2 4 ( ) 2( 1) () ( 1) a n nb b ax n a n nb b ax n  + −= +  + −= +

Check that since 2 ( 1)a n b+ , then 2 0nb b a+ −  For 22 00( 1) 2 0n x a x a nb+ − + −  22 0 ( ) ( ) 11 a n nb b a a n nb b a xnn− + − + + − ++ Where 2()M n nb b a= + − 6 Solutions (a) 61 4 (6 1) 5 19a = + −  = For the 6th row, the common difference is 3 (6 1) 2 13+ −  = . Hence, 66 19 (6 1) 13 84a = + −  = . (b) 1 4 3( 1) 3 1ma m m= + − = + For the thm row, the common difference is 3 2( 1) 2 1mm+ − = + . Hence, 3 1 ( 1)(2 1) 3 1 2 2 1 2 mna m n m m mn n m mn m n = + + − + = + + + − − = + + (c) 2 1 4 2 2 1 (2 1)(2 1) mna mn n m mn + = + + + = + + The necessary and sufficient condition for a number N to appear more than once in the array is that 2N+1 has more than one factorisation into two odd numbers greater than 1. For a number N in the array, if 2 1 (2 1)(2 1) (2 1)(2 1)N k h p q+ = + + = + + , where ( , )kh and ( , )pq are two different pairs of integers, then kh hk pq qpN a a a a= = = = , i.e. it appears more than twice. On t