1999 Solutions Fermat Contest (Grade 11)

Transcription

1 Canadian Mathematics Competition n activity of The Centre for Education in Mathematics and Computing, University of Waterloo, Waterloo, Ontario 999 s Fermat Contest (Grade ) for the wards 999 Waterloo Mathematics Foundation

2 999 Fermat s Part. The value of ( 9) is () 6 (B) 6 (C) (D) 8 (E) 9 ( ) = ( ) = NSWER: (E). Today is Wednesday. What day of the week will it be days from now? () Monday (B) Tuesday (C) Thursday (D) Friday (E) Saturday Since there are 7 days in a week it will be Wednesday in 98 days. In days it will be Friday. NSWER: (D). Six squares are drawn and shaded as shown. What fraction of the total area is shaded? () (B) (C) (D) (E) Out of a possible six squares, there is the equivalent of two shaded squares. Thus rd of the figure is shaded. NSWER: (B). Turning a screwdriver 9 will drive a screw mm deeper into a piece of wood. How many complete revolutions are needed to drive the screw 6 mm into the wood? () (B) (C) 6 (D) 9 (E) One complete revolution of the screw driver, 6, will drive it mm deeper into the wood. In order for the screw to go 6 mm into the wood it will take three revolutions. NSWER: () ( ) = is. value of x such that x () (B) (C) (D) (E)

3 999 Fermat s Since ( ) =, x = or x =. NWSER: (E) 6. The number which is 6 less than twice the square of is () 6 (B) (C) 6 (D) 8 (E) NSWER: (C) ( ) = 7. The Partridge family pays each of their five children a weekly allowance. The average allowance for each of the three younger children is $8. The two older children each receive an average allowance of $. The total amount of allowance money paid per week is () $ (B) $. (C) $ (D) $ (E) $ Solutiion The total paid out was, $ 8 + $ = $. NSWER: () 8. The time on a digital clock is :. How many minutes will pass before the clock next shows a time with all digits identical? () 7 (B) 7 (C) (D) 6 (E) 6 The digits on the clock will next be identical at :. This represents a time difference of 6 minutes. (Notice that times like 6:66, 7:77 etc. are not possible.) NSWER: (D) 9. In an election, Harold received 6% of the votes and Jacquie received all the rest. If Harold won by votes, how many people voted? () (B) 6 (C) 7 (D) (E) If Harold received 6% of the votes this implies that Jacquie received % of the total number of votes. The difference between them, %, represents votes. Therefore, the total number of votes cast was =. NSWER: (E). If x and y are each chosen from the set {,,,, }, the largest possible value of x y + is y x () (B) (C) (D) (E) The best strategy is to choose the largest value and the smallest so that, x >, is as large as possible. y

4 999 Fermat s When we consider the reciprocal, y, this will always produce a number less than and will be of little x consequence in our final consideration. The best choices, then, are x = and y = and x y + y x becomes + =. NSWER: (C) Part B. In Circle Land, the numbers 7 and are shown in the following way: 7 7 In Circle Land, what number does the following diagram represent? () (B) (C) (D) (E) = = = = = = The required number is + + =. Since there are four circles around the this corresponds to =. The corresponds to a in the units digit which leads to as the only correct possibility. NSWER: ()

5 999 Fermat s. The area of BC is 6 square units. If BD = 8 units and DC = units, the area (in square units) of BD is () (B) (C) 8 (D) 6 (E) 6 From, draw a line perpendicular to BC to meet BC at E. Thus the line segment E which is labelled as h is the height of BD and BC. Since the heights of the two triangles are equal, their areas are then proportionate to their bases. If the area of BC is 6, then the area of BD is 8 6 =. B 8 D C h B 8 E D C NSWER: (). Crippin wrote four tests each with a maximum possible mark of. The average mark he obtained on these tests was 88. What is the lowest score he could have achieved on one of these tests? () 88 (B) (C) (D) (E) If the average score of four tests was 88, this implies that a total of 88 or marks were obtained. The lowest mark would be obtained if Crippin had three marks of and one mark of. NSWER: (C). Three squares have dimensions as indicated in the diagram. What is the area of the shaded quadrilateral? () (B) 9 (C) (D) (E) In the first solution, we use similar triangles. We start by labelling the diagram as shown. The objective in this question is to calculate the lengths EB and FC which will allow us to calculate the area of EB and FC. We first note that FC and GD are similar and that, C FC D = GD = =. Therefore, FC = GD = ( )=. E B F C G D

6 999 Fermat s 6 Using the same reasoning, EB and FC are also similar triangles meaning that, EB FC =.. Thus, EB = ( ) = We find the required area to be area FC area EB = ( )( ) ( )( ) =. We start by putting the information on a coordinate axes and labelling as shown. The line containing OD has equation y x while x = and x = contains E and = BF. Solving the systems y= x, x = and y= x, x = gives the coordinates of E to be (, ) and F to be (, ). This makes E = and BF = which now leads to exactly the same answer as in solution. y x = x= E O (, ) B(, ) C(, ) F NSWER: () D(, ). If ( a+ b+ c+ d+ e+ f + g+ h+ i) is expanded and simplified, how many different terms are in the final answer? () 6 (B) 9 (C) (D) 8 (E) 7 Bracket Bracket ( a+ b+ c+ d+ e+ f + g+ h+ i) ( a+ b+ c+ d+ e+ f + g+ h+ i) If we wish to determine how many different terms can be produced we begin by multiplying the a in bracket by each term in bracket. This calculation gives 9 different terms. We continue this process by now multiplying the b in bracket by the elements from b to i in bracket to give 8 different terms. We continue this process until we finally multiply the i in the first bracket by the i in the second bracket. ltogether we have, = different terms. NSWER: (C) 6. If px + y = 7 and x+ qy= represent the same straight line, then p equals () (B) 7 (C) (D) (E) 7 If we multiply the equation of the first line by and the second by 7 we obtain, px + y = and x+ 7qy=. Comparing coefficients gives, p = or p =. NSWER: (D)

7 999 Fermat s 7 7. In BC, C = B = and BC =. D is a point chosen on BC. From D, perpendiculars are drawn to meet C at E and B at F. DE + DF equals E F () (B) (C) (D) (E) C D B We start by drawing a line from that is perpendicular to the base CB. Since BC is isosceles, M is the midpoint of CB thus making CM = MB =. Using pythagoras in CM we find M to be =. E F C M D B Join to D. The area of BC is ( )( )= but it is also, ( ED)( )+ ( DF)( ) ( ) = ED + DF. Therefore, ED + DF = ( )=. E F C D B NSWER: (C) 8. The number of solutions ( PQ, ) of the equation P Q from to 9 inclusive, is Q = P P+ Q, where P and Q are integers PQ () (B) 8 (C) 6 (D) 7 (E) 8 If we simplify the rational expression on the left side of the equation and then factor the resulting numerator as a difference of squares we obtain, ( P Q) ( P+ Q). PQ The equation can now be written as, ( P Q) ( P+ Q) P+ Q = or P Q= ( PQ and P + Q ). PQ PQ The only integers that satisfy this are: (, ), (, ), (, ),..., ( 9, 8). Thus there are 8 possibilities. NSWER: (B)

8 999 Fermat s 8 9. Parallelogram BCD is made up of four equilateral triangles of side length. The length of diagonal C is B C () (B) 7 (C) (D) (E) D From C, we draw a line perpendicular to D extended so that they meet at point E as shown in the diagram. This construction makes CDE a 6 9 triangle with CDE = 6 and CD =. Thus CE = and DE =. Using pythagoras in CE, we have E = and CE =, C = ( ) + ( ) = 7. B C D E NSWER: (B). If a () =, a x x a = x a a a =, and a a n = a n, for n, x and x, then a 7 is (B) x (C) x (D) x x ( )( ) x x x x = = ( x) = = = x x x x x () x x = = = x x x x x x x = x ( ) = x (E) x Since a = a, we conclude a = a = a7 =... = an = a6. lso, a = a = a8 =... = an = a7 for n = 6. x Since a = x then a7 =. NSWER: (D) x x

9 999 Fermat s 9 Part C. How many integers can be expressed as a sum of three distinct numbers if chosen from the set {, 7,,,..., 6}? () (B) 7 (C) 6 (D) (E) Since each number is of the form + n, n =,,,...,, the sum of the three numbers will be of the form + k+ l+ m where k, l and m are chosen from {,,,..., }. So the question is equivalent to the easier question of, How many distinct integers can be formed by adding three numbers from, {,,,..., }? The smallest is + + = 6 and the largest is + + =. It is clearly possible to get every sum between 6 and by: (a) increasing the sum by one replacing a number with one that is larger or, (b) decreasing the sum by one by decreasing one of the addends by. Thus all the integers from 6 to inclusive can be formed. This is the same as asking, How many integers are there between and 7 inclusive? The answer, of course, is 7. NSWER: (B) ( )( + ) and x x c ( x m) x+ n. If x + ax+ 8 = x+ y x z 8 + = + between and, then the maximum value of ac is ( ), where y, z, m, and n are integers () (B) 6 (C) (D) 78 (E) 98 For the equation, x + ax+ 8 = ( x+ y) ( x+ z) we consider the possible factorizations of 8 which produce different values for a. The factorizations and possible values for a are listed in the table that follows: Possible Factorizations of 8 Possible Values for a 8, 8 9, 9, 6, 6 6, 6 9, 9, 6, 6 6 8, 6 8, For the equation, x 8x+ c= ( x+ m) ( x+ n), we list some of its possible factorizations and the related possible values of c. Possible Factorizations Related Values of c ( x 9) ( x+ ) 9 = 9 x 8 x 8 = 9 ( )( + ) M M

10 999 Fermat s ( x 9) ( x+ ) = ( x 8) ( x+ ) 9 9 From these tables, we can see that the maximum value of ac is 9 9 = 98. NSWER: (E) x + x 6 ( ) =. The sum of all values of x that satisfy the equation x x+ is () (B) (C) (D) (E) 6 We consider the solution in three cases. Case It is possible for the base to be. Therefore, x x+ = x x+ = ( x )( x )= Therefore x = or x =. Both these values are acceptable for x. Case It is possible that the exponent be. Therefore, x + x 6= ( x+ )( x 6)= x = or x = 6 Note: It is easy to verify that neither x = nor x = 6 is a zero of x the indeterminate form does not occur. Case It is possible that the base is and the exponent is even. Therefore, x x+ = but x + x 6 must also be even. x x+ = x+, so that x x+ 6= ( x )( x )= x = or x = If x =, then x x 6 is even, so x = is a solution. If x =, then x x 6 is odd, so x = is not a solution. Therefore the sum of the solutions is =. NSWER: (B)

11 999 Fermat s. Two circles C and C touch each other externally and the line l is a common tangent. The line m is parallel to l and touches the two circles C and C. The three circles are mutually tangent. If the radius of C is 9 and the radius of C is, what is the radius of C? (). (B) (C) 8 (D) (E) 7 C C C m l We start by joining the centres of the circles to form CCC. (The lines joining the centres pass through the corresponding points of tangency.) Secondly, we construct the rectangle BC D as shown in the diagram. If the radius of the circle with centre C is r we see that: CC = r+, CC = r+ and CC =. 9 C D C C B m l We now label lengths on the rectangle in the way noted in the diagram. r C r 9 C r + r + 9 B r D C To understand this labelling, look for example at CD. The radius of the large circle is r and the radius of the circle with centre C is 9. The length CD is then r 9. This same kind of reasoning can be applied to both C and BC. Using Pythagoras we can now derive the following: In C C, C = ( r+) ( r ) = 6r. Therefore C= r. ( ) = +9 9 ( ) ( ) In DC C, DC r r = 6r. Therefore DC = r. 6 ( ) = ( ) In BC C, CB r = r + r. Therefore CB= r + r.

12 999 Fermat s In a rectangle opposite sides are equal, so: DC = C + CB or, 6 r = r + r + r r = r + r. Squaring gives, r = r + r r 8r = rr ( )= Therefore r = or r =. Since r >, r =. NSWER: (D). Given that n is an integer, for how many values of n is n n an integer? n n+ () 9 (B) 7 (C) 6 (D) (E) We start by dividing n n+ into n n. n n+ n n ) n 8n + 6 n This allows us to write the original expression in the following way, n n n n = + n n+ n n+ = + n n+. n + The original question comes down to the consideration of and when this expression is an n n+ integer. This rational expression can only assume integer values when, n+ n n+ (the numerator must be greater than the denominator) and when n + =. Or, n 6n 7 and n = or, ( n 7) ( n+ ) n 7. This means that we only have to consider values of n, n 7, n Z and n =. lso note that since n n+ =( n )( n ) we can remove n = and n = from consideration. We construct a table and check each value. n + n n ( )( ) n From this table we can see that there are just four acceptable values of n that produce an integer.

13 999 Fermat s n + Note also that would also be an integer if n + = and n n+. Thus n n+ n = is a fifth value since the denominator. NSWER: (E)