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