IMOK Maclaurin Paper 2014 1. What is the largest three-digit prime number whose digits, and are different prime numbers? We know that, and must be three of,, and. Let denote the largest of the three digits, the second largest and the smallest. If and then (else the three digits would sum to and the number would be divisibly by ), so. If and then (by similar reasoning; the digits would sum to ), so the two largest digits cannot be and. If, we must have and. So there are only two possible choices for the three digits, namely { } and { }. Any number made up of one of these two sets of digits contains both a and a. Since all 3-digit numbers which end in a are divisible by (and hence not prime) and all 3-digit numbers which end in a are divisible by (and hence not prime), and must take the values of and in some order, and must take the remaining value (either or ). So the only numbers which remain to check are (in descending order): ; ; ;. ; is prime 1 ; is prime 2 ; So and are the only two numbers with the desired properties and of those is the largest. 1 Once we observe that, it is sufficient to check divisibility by and, to test whether is prime, which is a simple exercise. 2 Since, we now only need check for divisibility by and, which is an even simpler exercise. 1
IMOK Maclaurin 2014 s 2. Nine buns cost pence and 13 buns cost pence, where and. What is the cost of a bun? Let denote the cost of a bun in pence. Then must be an integer. Then we have and. Simplifying each of these gives: and. Taking the stricter restriction for the upper and lower bounds on from these inequalities gives:, which, since must be a whole number of pence, gives p 2
IMOK Maclaurin 2014 s 3. A regular hexagon, with sides of length 2cm, is cut into two pieces by a straight line parallel to one of its sides. The ratio of the area of the smaller piece to the area of the larger piece is 1:5. What is the length of the cut? We can assume, without loss of generality, that the hexagon is oriented as below and that the cut is horizontal, with the smaller area above and the larger area below (if it is not then we can rotate it until it is): The area of the whole hexagon is ( ) So the smaller area (above the cut) is.. Consider just the trapezium above the cut: cm cm cm ( )cm Its top is just one side of the hexagon and so is cm in length. If its height is cm, then its base is ( ( )) Then its area,, is: cm. ( ) ( ) ( ) So, i.e., so ( ) ( ) So the cut has length ( ) 3
IMOK Maclaurin 2014 s 4. In the diagram, is the tangent at to the circle, and, an are all right angles. Prove that. Let. Then. Then, by the Alternate Segment Theorem,, and so. Hence triangles and are similar. Now let. Then. Then, by the Alternate Segment Theorem,, and so. Hence triangles and are similar. Since the triangles are similar, their side ratios are the same: Then i.e. and 4
IMOK Maclaurin 2014 s 5. Kim and Oli played nine games of chess, playing alternately with the white and black pieces. Exactly five games were won by whoever was playing with the black pieces, Kim won exactly six games, and no game was drawn. With which colour pieces did Kim play the first game? Let us assume that Kim played with the black pieces in the first game. Since Kim only played five games as black and four as white, she must have won at least one game playing as white and two games playing as black. If we assume (without loss of generality) that these were the final three games, we can reduce the problem to consider only the first six games (where Kim and Oli each won three times and the player with the black pieces won three times also). Since there are two players, and three of these six games were won by the player using the black pieces, one of the players must have won more than once playing with the black pieces (by the pigeonhole principle). If this player won three games as black, then they must have lost three times playing as white (meaning that their opponent also won three times playing as black), and the player with the black pieces won six times, which is too many. If this player won twice as black, then they won once as white and lost twice as white. This means that their opponent also won twice playing as black, which means that the player with the black pieces won four times, which is again too many. We have now reached a contradiction, and so our original assumption, that Kim played as black in the first game, must be incorrect. Thus Kim played the first match with the white pieces. A quick check shows that the required conditions are possible if Kim played the first game as white: if Kim wins the first six games (three with each colour) and then Oli wins the remaining three (two as white and one as black), then the player using the black pieces won five of the nine matches. 5
IMOK Maclaurin 2014 s 6. The T-tetromino is the shape made by joining four squares edge to edge, as shown. The rectangle has dimensions, where and are integers. The expression can be tiled by means that can be covered exactly with identical copies of without gaps or overlaps. (a) Prove that can be tiled by when both and are even. (b) Prove that cannot be tiled by when both and are odd. (a) If both and are both even, let and for integers and. Then has dimensions and can consequently be split up into squares, each with dimensions. A square can be tiled by (see below), and this tiling could be repeated times in order to tile. (b) If and are both odd, let and for integers and. Let be coloured as a chessboard, with alternate squares white and black. Then contains ( ) ( ) Clearly, must also contain ( ) white squares. ( ) Any T-tetromino must cover three squares within of one colour and one square of the other colour. Since there are equal numbers of black and white squares, the number of tetrominoes which cover three black squares and one w yp 1 m m m f m w ch cover three white yp 2 m Since there is an equal number of tetrominoes of each type, we can consider each tetromino as half of a pair (where a pair is one tetromino of each type). Clearly, each pair of tetrominoes covers 4 white squares and 4 black squares. Consequently, the number of black squares covered by a covering of is, where is an integer. Since there are ( ) black squares, the number of black squares is not a multiple of four, and so such a covering cannot exist. Hence cannot be tiled by. 6