NJC H3 Math 2024 Prelim Solutions

2024 NJC H3 Math Prelim Solutions 1 ( ) ( )( )( ) ( )( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 32 32 2 2 p x x a x b x c x d x e x f x a b c x ab bc ca x abc x d e f x de ef fd x def a b c d e f x ab bc ca de ef fd x abc def Sx ab bc ca de ef fd x abc def = + + + − − − − = + + + + + + + − − + + + + + − = + + + + + + + + − − − ++ = + + + − − − + + Since |S abc def+ and |S ab bc ca de ef fd+ + − − − , ( )|pSx for any integer x. In particular, ( )|pSd . Now, ( ) ( )( )( )| p |S d S d a d b d c + + + . If S is prime, by Euclid’s lemma, |S d a+ or |S d b+ or |S d c+ . WLOG, suppose that |S d a+ . Then, S d a+ as both S and da+ are positive. However, we now have S a b c d e f a d S= + + + + +  +  , which is a contradiction. 2(a) Out of (28 + 2) symbols in a line, choose 28 to be balls and 2 to be dividers. 28 2 4352 + = 2(b) Let 11 12yx=− . The problem is equivalent to finding the number of solutions to the equation 1 2 3 28 12y x x+ + = − where 1y , 2x and 3x are non-negative integers. Hence, 28 12 2 1532 −+ = 2(c) Let 11iiyx=− for 1,2,3i= . Then 1 2 3 28x x x+ + = becomes 1 2 3 5y y y+ + = where iy is non-negative for 1,2,3i= . 52 212 + = .

2024 NJC H3 Math Prelim Solutions 2(d) 12 kx x x n+ + + = ------- (*) Let S be the number of solutions to (*) where 12, , , kx x x are non-negative integers less than r. Generalising from (c), let 1iiy r x= − − for 1, 2, ,ik= . Then (*) becomes ( )12 1ky y y k r n+ + + = − − where iy is non-negative for 1, 2, ,ik= . Thus, ( ) ( ) 111 11 kr nk r n kS kk −− − − + −  ==  −−  . -----(1) Let iX be the set of solutions to (*) where 12, , , kx x x are non -negative integers and ixr for some 1, 2, ,ik= . Using the Principle of Inclusion and Exclusion, ( ) 1 2 1 2 3 1 2 1 2 31 1 1 1 2 3.... 1 ... , k o i j j j j j i j j k j j j k k k S S X X X X X X X X X X =        = − +  −   + + −        where oS is the number of solutions to (*) where 12, , , kx x x are non-negative integers. Generalising from (a), 1 .1 o nkS k +−= − Generalising from (b), 1 1 i n r kX k − + −= − for i = 1,2,…, k

2024 NJC H3 Math Prelim Solutions To find 12 , mj j jX X X   12where 1 ... , mj j j k     let iijjz x r=− if ijxr  . Then (*) becomes 1 2 1 remaining values of m m k ji j j j j j x z z z x x n mr + + + + + + + = − where all terms on the left are non-negative integers. Note that 12 0 mj j jX X X   = if 0 i.e. nn mr m r−   . [As an illustration, let’s consider finding 23XX for the equation in 2(c). Let 22 12zx=− and 33 12.zx=− Then 1 2 3 28x x x+ + = becomes 2 3 1 28 2(12) 4.z z x+ + = − = ] Hence, 12 1 if ,1 0 otherwise. mj j jX X X n mr k nmk r     − + − = −  Thus, by the Principle of Inclusion and Exclusion, ( ) ( ) ( ) 0 1 1 2 1 3 1 1 1 1 2 1 3 1 1.... 1 , where 1 112 1 m n r m m S n k k n r k k n r k k n r k k k k k k n mr k nmmk r k n mr k mk   = + − − + − − + − − + −          = − + −          − − − −           − + −   + + − =    −    − + −  = − −−−−−   −   Hence, by (1) and (2), ( ) 0 111 11 n r m m k n mr k kr n m k k   = − + − − −    −=     −−     .

2024 NJC H3 Math Prelim Solutions 3(i) ( )( ) ( ) 4 2 2 22 2 2 4 2 2 2 1 1 1 1 21 u u Cu u Cu u C u u u C u + = + + − + = + − = + + − Hence 22 0 2CC− =  = ( )( ) ( )( ) 22 22 4 2 1 2 1 2 1 2 1 1 Au B Au B u u u u Au B u u Au B u u u +− − + + − + + − + − − + + = + In numerator: Constant term: 11 2B B B+ =  = Coefficient of 2u : ( )2 2 0 21 2 2 2 2 A B A B BA − + − − = == Hence 4 22 1 1 1 1 1 222 2 2 2 1 2 1 2 1 uu u u u u u +− =−+ + + − + . 3(ii) 2 24 4 tan d2 sec 1d d1 d2 ux uu x ux uu xu = = = + +=

2024 NJC H3 Math Prelim Solutions ( ) ( ) ( ) 4 22 22 22 22 22 1 1 ddd dtan 2 d1 11 11 22 d 2 1 2 1 1 1 1 12 2 2 2 222 2 2 2 d 2 1 2 1 1 2 2 2 2 d 2 2 2 1 2 1 1 1 1 d2 2 1 2 1 1 ln 2 1 ln 2 1 22 11 2 1 2 xxu uux uu uu u u u u u uu u u u u u uu u u u u u u u u u u u u u u u = = + +− =− + + − + + + − − =− + + − + +−=− + + − + ++ + + − + = + + − − + +  +              ( ) ( ) 22 2 2 11 2 2 11 1 d 1 1 1 22 2 1 2 1ln 2 2 2 1 11 2 22tan tan 112 22 1 2 1ln1 2 21 2 tan 2 1 tan 2 1 1 tan 2 tan 1ln1 2 tan 2 tan 1 2 t u u uu uu uu C uu uu C uu xx xx −− −− + + − +    ++= −+     +−     + + +                  ++ −+=+  + + + − ++ −+= +    ( ) ( ) 11an 2 tan 1 tan 2 tan 1 C xx−−    +  + + −

2024 NJC H3 Math Prelim Solutions 4(a) (i) ( ) 2 1 2 2 21 ( 1) (2 1)! (2 1)! 1 2 (2 1) 1 2 (2 1) n n n x nn x n n n n nn nn + +−= + += + += + For all positive integers n, 1.n Thus, ( ) 2 211 2 4n+  = and 2 (2 1) 2 3 6nn +   = So 1 42 (shown)63 n n x x + = 4(a) (ii) 21 1 2 1 2 2 2 3 3 3 n n n nx x x x − −−              . Thus, ( ) 1 1 11 2 3 1 2 3 11 2 3 1 1 23 1 3 as ,3 nNN n nn N N xx x x x N − ==   −= −  = − → →   which is a finite number, hence the series 1 n n x  =  converges. 4(a) (iii) Since 11, 1 0 0. 2 lll − −    Take 1 .2 lk −= Then we can find a sufficiently large N such that if n > N, then 1 11 ,22 n n v lll k lv + −+ + = + = which is a positive number less than 1. Let 1.2 lr += Then for all positive integers m, 21 1 1 1 .mNm N m N m N m N Nm v r v rv r v r vv ++ + + + + − + +     

2024 NJC H3 Math Prelim Solutions Hence, for large integers M (larger than N), 1 1 1 1 11 1 11 1 1 1 1 1 1 as , since 0 1.1 M N M n n n n n n N NM m nN n n N NM m nN n n N MNN nN n N N n n v v v v r v v v r rvv r vv M r r = = = + + = = + + = = + − + = + = =+ + =+  −=+  − → + →  −        Since 1 M n n v =  is bounded above by a finite value for all positive integers N and the terms of the series are all positive, the series 1 n n v  =  must be convergent. 4(b) 1 1 1 12 1 1 1 1 1 1 1e e 1 (1) 1! 2! 3! 4! 5! 6! 7! 1 1 1 1 1 1 1e 1 (2) 1! 2! 3! 4! 5! 6! 7! 1 1 1(1) (2) : e e 2 1 3! 5! 7! 1 1 1 e e e 11 3! 5! 7! 2 2e − − − = = + + + + + + + + = − + − + − + − + − − = + + + + −−+ + + + = =

2024 NJC H3 Math Prelim Solutions 5(i) (a) Partition the set  1, 2,3,...,2m to m sets as follows:        1, 2 , 3, 4 , 5,6 ,..., 2 1, 2mm− . It can be seen that both elements in each set differ by 1 and are therefore coprime. By the Pigeonhole principle, since 1m+ integers are chosen, at least one of these sets will have both its elements chosen, which gives rise to the required pair of coprime integers. 5(i) (b) The largest odd divisor of each of the 1m+ integers can only possibly be from the set  1,3,5,..., 2 1m− which has m elements. By the Pigeonhole Principle, at least two of these 1m+ integers will share the same largest odd divisor, call it d. Thus, these two integers are of the form 2i d and 2 j d where ,ij + . Regardless of the relative values of i and j, one will divide the other. 5(ii) Let a b be an irreducible fraction in I with 1 bn . Let c kb , with k + and 1 kb n , be an irreducible fraction distinct from a b . Then, 11c a c ak kb b kb kb n −− =   . Thus, c Ikb as I is an open interval of length of 1 n . 5(iii) From (ii), it can be seen that every irreducible fraction in I with denominator at least 1 and at most n will have different denominators such that no two denominators have the property that one divides the other. If n is even, i.e., 2nm= , by (i)(b), there can be at most 2 nm= such irreducible fractions in I. If n is odd, i.e., 21nm=+ , by the given result, there can be at most 2 1 1 11 2 2 2 m n nm + + ++ = =  such irreducible fractions in I. Therefore, I contains at