UK I NTERMEITE MTHEMTIL OLYMPI ayley Question Papers and Solutions 2008 to 2010 Organised by the United Kingdom Mathematics Trust
i UKMT UKMT UKMT UK Intermediate Mathematical Olympiad 2008 to 2010 ayley Question Papers and Solutions Organised by the United Kingdom Mathematics Trust ontents ackground ii Rules and Guidelines 1 2008 paper 3 2009 paper 4 2010 paper 5 2008 solutions 6 2009 solutions 12 2010 solutions 18 UKMT 2011
ii ackground The Intermediate Mathematical Olympiad and Kangaroo (IMOK) are the follow-up competitions for pupils who do extremely well in the UKMT Intermediate Mathematical hallenge (about 1 in 200 are invited to take part). The IMOK was established in 2003. There are three written papers (ayley, Hamilton, Maclaurin) and two multiple-choice papers (the Pink and Grey Kangaroo). The written papers each take two hours and contain six questions. oth Kangaroo papers are one hour long and contain 25 questions. The ayley paper is for pupils in: Y9 or below (England and Wales); S2 or below (Scotland); School Year 10 or below (Northern Ireland).
1 The United Kingdom Mathematics Trust Intermediate Mathematical Olympiad and Kangaroo (IMOK) Olympiad ayley Paper ll candidates must be in School Year 9 or below (England and Wales), S2 or below (Scotland), or School Year 10 or below (Northern Ireland). RE THESE INSTRUTIONS REFULLY EFORE STRTING 1. Time allowed: 2 hours. 2. The use of calculators, protractors and squared paper is forbidden. Rulers and compasses may be used. 3. Solutions must be written neatly on 4 paper. Sheets must be STPLE together in the top left corner with the over Sheet on top. 4. Start each question on a fresh 4 sheet. You may wish to work in rough first, then set out your final solution with clear explanations and proofs. o not hand in rough work. 5. nswers must be FULLY SIMPLIFIE, and EXT. They may contain symbols such as π, fractions, or square roots, if appropriate, but NOT decimal approximations. 6. Give full written solutions, including mathematical reasons as to why your method is correct. Just stating an answer, even a correct one, will earn you very few marks; also, incomplete or poorly presented solutions will not receive full marks. 7. These problems are meant to be challenging! The earlier questions tend to be easier; the last two questions are the most demanding. o not hurry, but spend time working carefully on one question before attempting another. Try to finish whole questions even if you cannot do many: you will have done well if you hand in full solutions to two or more questions. O NOT OPEN THE PPER UNTIL INSTRUTE Y THE INVIGILTOR TO O SO! The United Kingdom Mathematics Trust is a Registered harity. Enquiries should be sent to: Maths hallenges Office, School of Maths Satellite, University of Leeds, Leeds, LS2 9JT. (Tel. 0113 343 2339) http://www.ukmt.org.uk
2 dvice to candidates o not hurry, but spend time working carefully on one question before attempting another. Try to finish whole questions even if you cannot do many. You will have done well if you hand in full solutions to two or more questions. nswers must be FULLY SIMPLIFIE, and EXT. They may contain symbols such as π, fractions, or square roots, if appropriate, but NOT decimal approximations. Give full written solutions, including mathematical reasons as to why your method is correct. Just stating an answer, even a correct one, will earn you very few marks. Incomplete or poorly presented solutions will not receive full marks. o not hand in rough work.
2008 3 1. How many four-digit multiples of 9 consist of four different odd digits? 2. hexagon is made by cutting a small equilateral triangle from each corner of a larger equilateral triangle. The sides of the smaller triangles have lengths 1, 2 and 3 units. The lengths of the perimeters of the hexagon and the original triangle are in the ratio 5 : 7. What fraction of the area of the original triangle remains? 3. In the rectangle the midpoint of is M and : = 2 : 1. The point X is such that triangle MX is equilateral, with X and lying on opposite sides of the line M. Find the value of X. 4. The number N is the product of the first 99 positive integers. The number M is the product of the first 99 positive integers after each has been reversed. That is, for example, the reverse of 8 is 8; of 17 is 71; and of 20 is 02. Find the exact value of N M. 5. kite has sides and of length 25 cm and sides and of length 39 cm. The perpendicular distance from to is 24 cm. The perpendicular distance from to is h cm. Find the value of h. 39 cm 39 cm h cm 24 cm 25cm 25cm 6. regular tetrahedron has edges of length 2 units. The midpoint of the edge is M and the midpoint of the edge is N. Find the exact length of the segment MN.
4 2009 1. n aquarium contains 280 tropical fish of various kinds. If 60 more clownfish were added to the aquarium, the proportion of clownfish would be doubled. How many clownfish are in the aquarium? 2. The boundary of the shaded figure consists of four semicircular arcs whose radii are all different. The centre of each arc lies on the line, which is 10 cm long. What is the length of the perimeter of the figure? 3. Two different rectangles are placed together, edge-to-edge, to form a large rectangle. 2 The length of the perimeter of the large rectangle is 3 of the total perimeter of the original two rectangles. Prove that the final rectangle is in fact a square. 4. In the rectangle, the side has length 2 and the side has length 1. Let the circle with centre and passing through meet at X. Find X (in degrees). 5. Two candles are the same height. The first takes 10 hours to burn completely whilst the second takes 8 hours to burn completely. oth candles are lit at midday. t what time is the height of the first candle twice the height of the second candle? 6. Teams,, and competed against each other once. The results table was as follows: Team Win raw Loss Goals for Goals against 3 0 0 5 1 1 1 1 2 2 0 2 1 5 6 0 1 2 3 6 (a) Find (with proof) which team won in each of the six matches. (b) Find (with proof) the scores in each of the six matches.
2010 5 1. The sum of three positive integers is 11 and the sum of the cubes of these numbers is 251. Find all such triples of numbers. 2. The diagram shows a square and an equilateral triangle E. The point F lies on so that E = EF. alculate the angle EF. E F 3. Find all possible solutions to the word sum on the right. Each letter stands for one of the digits 0 9 and has the same meaning each time it occurs. ifferent letters stand for different digits. No number starts with a zero. O + O E V E N 4. Walking at constant speeds, Eoin and his sister ngharad take 40 minutes and 60 minutes respectively to walk to the nearest town. Yesterday, Eoin left home 12 minutes after ngharad. How long was it before he caught up with her? 5. square sheet of paper is folded along FG, as shown, so that the corner is folded onto the midpoint M of. Prove that the sides of triangle GM have lengths in the ratio 3 : 4 : 5. M G F 6. qprime number is a positive integer which is the product of exactly two different primes, that is, one of the form q p, where q and p are prime and q p. What is the length of the longest possible sequence of consecutive integers all of which are qprime numbers?
6 2008 Solutions 1. How many four-digit multiples of 9 consist of four different odd digits? First solution There are five odd digits: 1, 3, 5, 7 and 9. The sum of the four smallest odd digits is 16 and the sum of the four largest is 24. Hence the digit sum of any four-digit number with different odd digits lies between 16 and 24, inclusive. However, the sum of the digits of a multiple of 9 is also a multiple of 9, and the only multiple of 9 between 16 and 24 is 18. Hence the sum of the four digits is 18. Now 1 + 3 + 5 + 9 = 18, so that the four digits can be 1, 3, 5 and 9. If 7 is one of the four digits then the sum of the other three is 11, which is impossible. So 7 cannot be one of the digits and therefore the four digits can only be 1, 3, 5 and 9. The number of arrangements of these four digits is 4 3 2 1 = 24. Hence there are 24 four-digit multiples of 9 that consist of four different odd digits. Second solution The sum of all five odd digits is 1 + 3 + 5 + 7 + 9 = 25. Subtracting 1, 3, 5, 7 and 9 in turn we get 24, 22, 20, 18 and 16, only one of which is a multiple of 9, namely 18 = 25 7. Since the sum of the digits of a multiple of 9 is also a multiple of 9, it follows that the four digits can only be 1, 3, 5 and 9. The number of arrangements of these four digits is 4 3 2 1 = 24. Hence there are 24 four-digit multiples of 9 that consist of four different odd digits. 2. hexagon is made by cutting a small equilateral triangle from each corner of a larger equilateral triangle. The sides of the smaller triangles have lengths 1, 2 and 3 units. The lengths of the perimeters of the hexagon and the original triangle are in the ratio 5 : 7. What fraction of the area of the original triangle remains? First solution Let the side length of the large equilateral triangle be x units; this triangle therefore has a perimeter of length 3x units. Now consider the hexagon, which has sides of lengths 1, x 3, 2, x 5, 3 and x 4 units. Hence the hexagon has perimeter length 3x 6 units. Since the ratio of the perimeter lengths of the hexagon and the large triangle is 5 : 7, we have 3x 6 3x Rearranging and solving for x we obtain x = 7. = 5 7. ( )
7 Now, in order to find the area of the large equilateral triangle, we determine the height units using Pythagoras' theorem: h 2 = 7 2 ( 7 2) 2 = 49 (1 1 4) = 49 3 4. Hence h = 49 3 4 = 7 3 2. Therefore the area of the large equilateral triangle is 1 2 7 7 3 = 49 3 2 4. We may find the areas of the three small equilateral triangles in a similar way. These areas are 3 4, 4 3 4 and 9 3 4. The area of the hexagon is the area of the large equilateral triangle minus the areas of the three small equilateral triangles, that is, 49 3 3 4 ( 4 + 4 3 + 9 3 4 4 ) = 35 3 4. Finally, the fraction of the original equilateral triangle remaining is 35 3 4 49 3 4 = 5 7. Second solution Having established that the large triangle has sides of length 7 (equation (*) in the solution above), we may proceed as follows: The four equilateral triangles in the problem are similar. Now the ratio of the areas of similar figures is equal to the ratio of the squares of their sides. Hence the four triangles have areas in the ratio 1 2 : 2 2 : 3 2 : 7 2 = 1 : 4 : 9 : 49. Hence the ratio of the areas of the hexagon and the large triangle is 49 (1 + 4 + 9) : 49 = 35 : 49 = 5 : 7. This may be illustrated by dividing the large triangle into 49 small triangles, as shown. h Note: The observant reader will have noticed that the answer to this problem is surprising: the ratio of the areas is the same as the ratio of the perimeters. There is no reason to expect this to happen.
8 3. In the rectangle the midpoint of is M and : = 2 : 1. The point X is such that triangle MX is equilateral, with X and lying on opposite sides of the line M. Find the value of X. Solution The key to this solution is to draw M and consider triangle MX. We are given that is a rectangle, so that = and M = 90 = M. We are also given that = 2 and that M is the midpoint of. Therefore = M = M =. X It follows that triangles M and M are congruent (SS) and we deduce that M = M. ut triangle MX is equilateral, so MX = M and hence MX = M. In other words, triangle MX is isosceles. Now consider the angles at M. 1. Triangle M is right-angled with M = 90. It is also isosceles, so M = M = 45, since the angle sum is 180. 2. Similarly, from triangle M, M = 45. 3. Finally, because triangle MX is equilateral, MX = 60. Hence M MX = 180 45 45 60 = 30 since angles on a straight line add up to 180. Lastly, we consider the angles at. We know that triangle XM is isosceles and that MX = 30. Hence each base angle 1 is 2 (180 30 ) = 75 ; in particular, XM = 75. lso, is a rectangle, so = 90, and triangle M is right-angled and isosceles, so M = 45. Therefore M = M = 90 45 = 45. We can now calculate the value of X. We have X = XM M = 75 45 = 30.
9 4. The number N is the product of the first 99 positive integers. The number M is the product of the first 99 positive integers after each has been reversed. That is, for example, the reverse of 8 is 8; of 17 is 71; and of 20 is 02. Find the exact value of N M. First solution From the given definition we have N = (1 2 9) 10 (11 12 19) 20 90 (91 99), which rearranges to lso N = (1 2 9) (11 12 19) (91 99) (10 20 90). M = (1 2 9) 01 (11 21 91) 02 09 (19 99) which rearranges to M = (1 2 9) (11 12 19) (91 99) (01 02 09) = (1 2 9) (11 12 19) (91 99) (1 2 9). omparing these arrangements for M and N, we see that M has the same terms as N except that the product 10 20 90 is replaced by the product 1 2 9. Thus when we divide N by M all the common terms cancel and we are left with N M = 10 20 90 1 2 9 = 10 9. Second solution We may place the numbers from 1 to 99 into three categories, determined by how they are transformed when they are reversed: 1. single digit numbers a are unchanged; 2. a two-digit number ab, where neither a nor b is zero, is transformed to the twodigit number ba ; and 3. a multiple of 10 such as a0 is transformed to 0a = a, a single-digit number. Thus there is a correspondence between the factors in N and M, as shown in the table: N M a a aa aa ab and ba ba and ab a0 a Single-digit numbers are unchanged; two-digit numbers with a repeated digit are unchanged; pairs of two-digit numbers, with different digits and neither digit zero, are unchanged as a pair; the multiples of 10 in N are replaced by single-digit numbers in M. Thus when we divide N by M all the identical factors cancel and we are left with N M = 10 20 90 1 2 9 = 10 9.
10 5. kite has sides and of length 25 cm and sides and of length 39 cm. The perpendicular distance from to is 24 cm. The perpendicular distance from to is h cm. Find the value of h. 39 cm 39 cm h cm 24cm 25 cm 25 cm First solution s shown in the figure below, let the perpendicular from to the line meet the line at the point X; let the perpendicular from to the line meet the line at the point Y and let the distance Y be y cm. 39 y 39 Y h y 25 24 X 25 onsidering triangle X and using Pythagoras' Theorem we obtain X = = 7 cm. Similarly, from triangle X we have = 25 2 24 2 cm 24 2 + (25 7) 2 cm = 30 cm. ( ) Now from triangles Y and Y, again by Pythagoras' theorem, we deduce that h 2 + (39 y) 2 = 39 2 and h 2 + y 2 = 30 2. (1) Subtract to get (39 y) 2 y 2 = 39 2 30 2, which simplifies to 78y = 900, 150 so that y =. 13 Finally, by substituting in equation (1), we find h = 360 13. Second solution nother solution uses the length of obtained in (*) above to find the area of isosceles triangle. Once the area is known the value of the height h may be found from area = 1 2 39 h. an you see how to find the area of triangle and so complete the solution? Note: Triangle Y is a 5, 12, 13 triangle.
6. regular tetrahedron has edges of length 2 units. The midpoint of the edge is M and the midpoint of the edge is N. Find the exact length of the segment MN. M 11 First solution We make use of the following result. Theorem (Median of isosceles triangle): The line joining the apex to the midpoint of the base of an isosceles triangle is perpendicular to the base. That is, in the following figure, PSR = 90. N R pplying the theorem to triangle, we find that M = 90. Similarly, in triangle, M = 90. Now applying Pythagoras' theorem to the triangles M and M we get M 2 = 2 M 2 = 2 2 1 2 and M 2 = 2 M 2 = 2 2 1 2. Hence M = 3 and M = 3, so triangle M is isosceles. Now apply the theorem to triangle M to obtain NM = 90. Then by Pythagoras' theorem in triangle NM Therefore MN = 2. MN 2 = M 2 N 2 = 3 1. Second solution tetrahedron may be formed by joining face diagonals of a cube, as shown below. Since the faces of the cube are congruent squares the face diagonals have equal length and so the tetrahedron is regular. Now M and N are midpoints of opposite edges of the tetrahedron. Therefore they are midpoints of opposite face diagonals of the cube, that is, centres of opposite faces of the cube. Hence MN = R. Letting the sides of the cube have length a, from Pythagoras' theorem in triangle R we get P so that 2 = R 2 + R 2 2 2 = a 2 + a 2 = 2a 2. Hence a = 2 and therefore MN = 2. S M P N S Q R Q
12 2009 Solutions 1. n aquarium contains 280 tropical fish of various kinds. If 60 more clownfish were added to the aquarium, the proportion of clownfish would be doubled. How many clownfish are in the aquarium? Solution Let there be x clownfish in the aquarium. If 60 clownfish are added there are x + 60 clownfish and 340 tropical fish in total. Since the proportion of clownfish is then doubled, we have 2 x 280 = x + 60 340. Multiplying both sides by 20 we get and hence It follows that x 7 = x + 60 17 17x = 7 (x + 60). x = 42 and thus there are 42 clownfish in the aquarium. 2. The boundary of the shaded figure consists of four semicircular arcs whose radii are all different. The centre of each arc lies on the line, which is 10 cm long. What is the length of the perimeter of the figure? Solution The centre of the large semicircular arc lies on, so we know that is a diameter of the large semicircle. ut is 10 cm long, so the radius of the large semicircle is 5 cm. Let the radii of the other three semicircles be r 1 cm, r 2 cm and r 3 cm. The centres of these arcs also lie on, so the sum of their diameters is equal to the length of. It follows that 2r 1 + 2r 2 + 2r 3 = 10 and hence r 1 + r 2 + r 3 = 5. Now the lengths, in cm, of the semicircular arcs are 5π, πr 1, πr 2 and πr 3. Therefore the perimeter of the figure has length, in cm, 5π + πr 1 + πr 2 + πr 3 = π (5 + r 1 + r 2 + r 3 ) Hence the perimeter of the figure has length = π (5 + 5) = 10π. 10π cm.
3. Two different rectangles are placed together, edge-to-edge, to form a large rectangle. 2 The length of the perimeter of the large rectangle is 3 of the total perimeter of the original two rectangles. 13 Prove that the final rectangle is in fact a square. First solution Since the smaller rectangles are placed together edge-to-edge, they have a side length in common. Let this side have length y and let the other sides have lengths x 1 and x 2 as shown. y x 1 x 2 The perimeters of the smaller rectangles are 2x 1 + 2y and 2x 2 + 2y, so the total perimeter of the two smaller rectangles is 2x 1 + 2x 2 + 4y. The perimeter of the large rectangle is 2 (x 1 + x 2 ) + 2y = 2x 1 + 2x 2 + 2y. 2 We are given that the length of the perimeter of the large rectangle is 3 of the total perimeter of the two original rectangles. Hence we may form the equation 2x 1 + 2x 2 + 2y = 2 3 (2x 1 + 2x 2 + 4y). We may simplify this equation by multiplying both sides by 3 and expanding the brackets, to obtain which simplifies to 6x 1 + 6x 2 + 6y = 4x 1 + 4x 2 + 8y, x 1 + x 2 = y. This means that the length and width of the large rectangle are the same. In other words, the rectangle is actually a square. Second solution The total perimeter length P of the original two rectangles is equal to the perimeter length of the large rectangle added to the lengths of the two edges which are joined together. 2 ut the perimeter length of the large rectangle is 3 P and hence the two edges which are 1 joined together have total length 3P. However, the two edges which are joined together are parallel to two sides of the large rectangle and have the same length as them. Hence these two sides of the large rectangle 1 have total length 3 P. Since the perimeter length of the large rectangle is 2 3 P, the other two sides of the large 1 rectangle also have total length 3 P. It follows that all the sides of the rectangle are equal in length, in other words, the rectangle is a square.
14 4. In the rectangle, the side has length 2 and the side has length 1. Let the circle with centre and passing through meet at X. Find X (in degrees). Solution We begin with a diagram showing the information given in the question. We have used the fact that is a rectangle, so that = = 1 and = = 2, and angles, and are right angles. 2 1 X Since X and are both radii of the circle, X also has length 1. This means that triangle X is isosceles and so X = X. Furthermore, since is a right angle, X and X are both equal to 45. From the fact that is a right angle, it follows that X = 90 45 = 45. We may use Pythagoras' theorem in triangle X to obtain X 2 = X 2 + 2 = 1 2 + 1 2 = 2 and so X = 2. We are given that also has length 2 and so triangle X is isosceles. This means that X and X are equal, and so each is equal to (180 45 ) 2 = 67 1 2. Lastly, we use the fact that is a right angle to conclude that X = 90 67 1 2 = 22 1 2.
15 5. Two candles are the same height. The first takes 10 hours to burn completely whilst the second takes 8 hours to burn completely. oth candles are lit at midday. t what time is the height of the first candle twice the height of the second candle? Solution h Let the initial height of each candle be h cm. In one hour the first candle will burn 10 cm h and the second candle will burn cm. Thus in hours, the candles will burn 8 t ht 10 cm and ht 8 cm, respectively. If both candles are lit at midday, then t hours after midday the heights of the first and second candles will be ( h 10) ht ( cm h ht 8 ) and cm, respectively. We are asked to find the time at which the height of the first candle is twice the height of the second candle. We therefore need to find the value of t such that h ht 10 ( = 2 h ht 8 ). We may divide every term by h, since we know that h is not zero, and expand the brackets to obtain the equation 1 t 10 = 2 t 4. Multiplying both sides by 20, we get and so 20 2t = 40 5t, t = 20 3 = 6 2 3. Hence the height of the first candle is twice that of the second after 6 hours and 40 minutes, in other words, this happens at 18:40.
16 6. Teams,, and competed against each other once. The results table was as follows: Team Win raw Loss Goals for Goals against 3 0 0 5 1 1 1 1 2 2 0 2 1 5 6 0 1 2 3 6 (a) Find (with proof) which team won in each of the six matches. (b) Find (with proof) the scores in each of the six matches. Solution (a) Team won all three games and so beat teams, and. Of the three games that team played, the one that was lost can only have been against team. Therefore team drew against teams and. If we consider the three games that team played, the game against team was lost, the game against team was a draw and so the remaining game, that team won, was against team. In summary: beat, beat, beat ; drew with, beat ; and drew with. (b) onsider the following table in which the rows give the number of goals scored for each team and the columns give the number of goals against each team. Goals against ll z + 1 5 Goals x t 2 for z x y 5 y 3 ll 1 2 6 6 15 We have let the number of goals scored by team against team be x, so that the number of goals scored by team against team is also x, since their match was a draw. Similarly, we have let the number of goals scored by team against team be y, so that this is also the number scored by team against team. Furthermore, we have let the number of goals scored by team against team be z, so that the number of goals scored by team against team is z + 1 since the difference between the number of goals scored and conceded by team is 1. Finally, we have let the number of goals scored by team against team be t. Then t is at least 1 since team beat team.
17 We observe that the row for now means that x + y + z = 5 (which agrees with the column for ). From the column for we see that z is at most 1, since the total in that column is 1. Similarly, from the row for, we see that y is at most 3, and from the row for we see that x is at most 1 since t is at least 1. ut we have x + y + z = 5, so that the only possibilities are x = 1, y = 3 and z = 1. It follows that t = 1. Therefore the table is: Goals against ll 2 5 Goals 1 1 2 for 1 1 3 5 3 3 ll 1 2 6 6 15 We may now complete the table by, for example, first noting that all other entries in the column for are 0, and then filling in the rows from the bottom. Goals against ll 1 2 2 5 Goals 0 1 1 2 for 1 1 3 5 0 0 3 3 ll 1 2 6 6 15 In summary, the scores in each match were as follows: beat 1 0 beat 2 1 beat 2 0 drew with 1 1 beat 1 0 drew with 3 3
18 2010 Solutions 1 The sum of three positive integers is 11 and the sum of the cubes of these numbers is 251. Find all such triples of numbers. Solution Let us calculate the first few cubes in order to see what the possibilities are: 1 3 = 1, 2 3 = 8, 3 3 = 27, 4 3 = 64, 5 3 = 125, 6 3 = 216 and 7 3 = 343. ( ) The sum of the cubes of the positive integers is 251, which is less than 343, hence none of the integers is greater than 6. Now 3 = 83 2 3 > 64 = 4 3, therefore at least one of the integers is 5 or more. 251 If one of the integers is 6, then the other two cubes add up to 251 6 3 = 251 216 = 35. From (*) above, 3 3 + 2 3 = 27 + 8 = 35 is the only possibility. lso, 6 + 3 + 2 = 11 so that 6, 3 and 2 is a possible triple of numbers. If one of the integers is 5, then the other two cubes add up to 251 5 3 = 251 125 = 126. From (*) above 5 3 + 1 3 = 125 + 1 = 126 is the only possibility. lso, 5 + 5 + 1 = 11 so that 5, 5 and 1 is a possible triple of numbers. Hence 2, 3, 6 and 1, 5, 5 are the triples of numbers satisfying the given conditions. 2 The diagram shows a square and an equilateral triangle E. The point F lies on so that E = EF. alculate the angle EF. E F Solution The diagram seems to include several isosceles triangles, and we solve the problem by proving this is the case. For example, since the square and the equilateral triangle E share the side, all their sides are the same length. That means the triangle E is isosceles. E 75 F Now, angle E is 90 60 = 30 (since it is the difference between the interior angle of a square and the 30 interior angle of an equilateral triangle). Hence angles 1 E and E are each 2 (180 30 ) = 75, because 60 they are the base angles of an isosceles triangle. We are also given that triangle EF is isosceles. Since we have worked out that angle FE = 75, we deduce that angle EF = 180 (2 75 ) = 30. Finally, we find that angle EF = E EF = 75 30 = 45.
3 Find all possible solutions to the word sum on the right. Each letter stands for one of the digits 0 9 and has the same meaning each time it occurs. ifferent letters stand for different digits. No number starts with a zero. O + O E V E N Solution Firstly, it is clear that the three-digit number O lies between 100 and 999. Therefore, since EVEN = 2 O, we have 200 < EVEN < 1998. Hence the first digit E of EVEN is 1 since it is a four-digit number. 19 We are left with the following problem: O + O 1 V 1 N Now the same numbers are added in the tens and units columns, but N 1, otherwise N and E would be equal. The only way for different totals to occur in these columns is for there to be a carry to the tens column, and the greatest possible carry is 1, so that N = 0. There are two possible digits that give N = 0, namely 0 and 5. ut 0 is already taken as the value of N, so that = 5. The problem is thus: O 5 5 + O 5 5 1 V 1 0 Now, the digit O has to be big enough to produce a carry, but cannot be 5, which is already taken as the value of. So the possibilities are 6 5 5 + 6 5 5 1 3 1 0 7 5 5 + 7 5 5 1 5 1 0 8 5 5 + 8 5 5 1 7 1 0 9 5 5 + 9 5 5 1 9 1 0 but the second and fourth of these are not allowed since V repeats a digit used for another letter. We are left with the two possibilities 6 5 5 + 6 5 5 1 3 1 0 and it is clear that both of these work. 8 5 5 + 8 5 5 1 7 1 0 4 Walking at constant speeds, Eoin and his sister ngharad take 40 minutes and 60 minutes respectively to walk to the nearest town. Yesterday, Eoin left home 12 minutes after ngharad. How long was it before he caught up with her? Solution Let the distance from home to town be km. Now in every minute Eoin travels onefortieth of the way to town: that is, a distance of. So after minutes, he has 40 km t travelled a distance t 40 km.
20 Similarly, in every minute ngharad travels one-sixtieth of the way to town: that is, a distance of. ut she has had 12 minutes extra walking time. So after Eoin has 60 km been walking for t minutes, she has been walking for t + 12 minutes and so has travelled a distance (t + 12) km. 60 We are asked how long Eoin has been walking when they meet. They meet when they have travelled equal distances, which is when t (t + 12) =. 40 60 We cancel the from each side and multiply both sides by 120 to obtain Simplifying, we get 120 t 40 = 120 t + 12 60. 3t = 2 (t + 12), which we solve to give t = 24. Thus Eoin catches up with ngharad after he has walked for 24 minutes. 5 square sheet of paper is folded along FG, as shown, so that the corner is folded onto the midpoint M of. Prove that the sides of triangle GM have lengths in the ratio 3 : 4 : 5. M G F Solution This problem does not give us units, and so we choose them so that the side length of the square is 2s. Since M is the midpoint of, we have M = s. Then we define x = G. Since = 2s, G = 2s x. ut, as GM is the image of G after folding, GM = 2s x too. s M s 2s x x G 2s x F Now Pythagoras' theorem for triangle MG gives us s 2 + x 2 = (2s x) 2. We multiply out to get s 2 + x 2 = 4s 2 4sx + x 2.
Eliminating the which has the solution x = 3 4s. x 2 terms and dividing by s (which is not zero), we obtain s = 4s 4x, 21 Thus the triangle GM has sides of length x = 3 4 s, s and 2s x = 5 4 s. Multiplying all the sides by 4, we get 3s, 4s and 5s, so the side lengths are in the ratio 3 : 4 : 5, as required. 6 qprime number is a positive integer which is the product of exactly two different primes, that is, one of the form q p, where q and p are prime and q p. What is the length of the longest possible sequence of consecutive integers all of which are qprime numbers? Solution To help to understand this problem, it is natural to test the first few numbers to see which small numbers are qprime, and which are not: 1 is not a qprime since it has no prime factors. 2 and 3 are not qprimes since they are prime. 4 is not qprime since it is 2 2. 5 is not qprime, since it is prime. 6 = 2 3 is the first qprime number. 7 is not qprime. 8 is not qprime, since it is 2 2 2. 9 is not, since it is 3 3. 10 = 2 5 is another qprime. 11 is not. 12 is not, since it is 2 2 3. Of course, we cannot prove a general result just by continuing the list, but it can guide us to a proof, such as the one that follows. We note that no multiple of 4 is ever qprime, since a multiple of 4 is a multiple of 2 2. This means that a string of consecutive qprime numbers can be of length at most three, because any sequence of four or more consecutive integers includes a multiple of 4. We are therefore led to ask whether any strings of three consecutive qprime numbers exist. We have looked as far as 12 and not found any, but we will continue searching, using the fact that none of the numbers is a multiple of 4: For (13, 14, 15), the number 13 is prime and so not qprime. For (17, 18, 19), 17 is not qprime (nor are the others). For (21, 22, 23), 23 is not qprime. For (25, 26, 27), 25 is not qprime (nor is 27). For (29, 30, 31), 29 is not qprime (nor are the others). For (33, 34, 35), all three are qprime (being 3 11, 2 17 and 5 7). So we have found a sequence of three consecutive qprimes, and have also proved that no sequence of four (or more) consecutive qprimes exists. Thus the longest possible sequence of consecutive integers all of which are qprime numbers has length 3.