Mathematical Challenge 1998-99

Problem Set 3 : Solutions

P1. If the "added" number is not 1, then the answer will be bigger if we interchange the positions of the added number and 1. For example, the answer for 31   52 + 4 is 1616, whilst the answer for 34   52 + 1 is 1769; and the answer for 452   3 + 1 is 1357 .

When the added number is 1, the numbers in the multiplication use each of 2, 3, 4 and 5 exactly once. There are two cases:
 
 

(1) a 3-digit number   a 1-digit number;

(2) a 2-digit number   a 2-digit number.
 
 

 For the largest answer, the digits in any 2- or 3- digit number involved are in descending order, reading from left to right. Therefore the multiplication part of the sum is whichever of the following has the biggest value:
  Case (1) 543    2 ; 542    3 ; 532    4 ; 432    5 .

Case (2) 54   32 ; 53   42 ; 52   43 .
 
 

 The values are:

Case (1) 1086 , 1626 , 2128 and 2160 respectively

Case (2) 1728 , 2226 , and 2236 respectively.

The biggest value for the multiplication is therefore 52   43 = 2236 , and so the biggest possible answer for a sum as defined in the question is 52   43 + 1 = 2237 .
 
 

P2 = J1 If the gardener had not planted any beds on the first day, he would have planted 12 + 24 + 36 + 48 = 120 in all. In fact, he planted 200. Since 200 - 120 = 80, he planted a further 80 / 5 = 16 each day. Therefore he planted 16 on the first day.
 

 

 
 
 
 
 
 

 P3 = J2 If I have only 5p, 2p and 1p coins and cannot make a total of exactly 10p, then the greatest sum of money I can have is 13p: one 5p coin and four 2p coins. Without the 5p coin, the greatest sum would be 9p. Similarly, if I have only 50p, 20p and 10p coins and cannot make a total of exactly 1.00, the greatest sum of money that I can have is 1.30: one 50p coin and four 20p coins. Without the 50p coin, the greatest sum would be 90p. The totals that I can make with one 50p coin and four 20p coins are 20p, 40p, 50p, 60p, 70p, 80p, 90p, 1.10 and 1.30. If I also have 5p, 2p and 1p coins in my purse which cannot make an exact total of 10p, I shall still not be able to make exactly 1.00. The greatest possible sum from these extra coins is 13p. Hence if I cannot make the exact 1.00 fare, I can have as much as 1.30 + 13p = 1.43 in my purse. However, there cannot be more than this, and so the largest sum of money I could have in my purse is 1.43.  

 
 
 
 
 
 

 J3. We know that X + Y = 99, that X is a multiple of 4 and that Y is a multiple of 7. Since X is positive, Y is less than 99. The possible values of Y are therefore 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91 and 98. Of these, each of 21, 42, 63 and 84 is 7 times the sum of its digits. The corresponding values of X, given by X + Y = 99, are 78, 57, 36 and 15, of which only 36 is a multiple of 4. We can verify that 36 is 4 times the sum of the digits, and so the required values are given by: X = 36 and Y = 63 .
 
 

J4 = M2 The answer is "No". The square of an even positive integer is always even, and so the final digit is even. For the odd positive integers, the final digit of the square is odd, but for squares greater than 10, the tens digit is always even.
 

  Proof. We can express any odd integer n greater than 10 in the form 10x + y, where x is a positive integer and y is one of 1, 3, 5, 7 and 9. Then n2 = 100x2 + 20xy + y2 , and y2 is one of 1, 9, 25, 49 and 81. Using 25 = 20 + 5, 49 = 2   20 + 9 and 81 = 4   20 + 1, we see that n2 is expressible as 20    a positive integer + one of 1, 5 and 9. The tens digit of n2 is therefore even.   (The last two digits of the squares of the odd integers 5, 7, 9, 11, 15, 17, 19, 21, 23 and 25 are 25, 49, 81, 21, 69, 25, 89, 61, 41, 29 and 25 (A)  respectively. The last two digits of the squares of the odd integers from 25 to 49 are these same numbers but in reverse order, with 09 and 01 added: 25, 29, 41, 61, 89, 25, 69, 21, 81, 49, 25, 09, and 01. (B) The squares of the odd integers from 51 to 99 inclusive are: 01, 09, 25, 49, 81, 21, 69, 25, 89, 61, 41, 29, 25, 29, 41, 61, 89, 25, 69, 21, 81, 49, 25, 09, 01   and for the squares of the subsequent odd integers, the same pattern is repeated indefinitely.)  

 
 
 
 
 
 

 M1 Divide 1,000,000 by 7. The quotient is 142,857 and the remainder is 1. Hence 999,999 = 7   142,857 and so there are 142,857 positive integers less than 1,000,000 which are divisible by 7. A number which is divisible by both 7 and 13 is divisibly by 7   13 = 91, since 7 and 13 have no common factors other than 1. Divide 999,999 by 91; we get 999,999 = 91   10,989. Hence there are 10,989 positive integers less than 1,000,000 which are divisible by 91.

It follows that there are 142,857 - 10,989 = 131,868 positive integers less than 1,000,000 which are divisible by 7 but not by 13.
 
 

M3 = S1 Since Rose and Sally were both walking at 3 mph and were walking towards one another, they were  mile apart when they had each covered  mile, which was after 8 minutes, and they met 2 minutes later. Hence Rex ran at 9 mph for 8 minutes and at 12 mph for 2 minutes, and so
 

  from the time Rex first saw Sally until Rose and Sally met, Rex covered  mile.  

 

M4 = S2
 

 

 
 
 
 

 In the diagram, ~ represents the nearer bank, G the gate and C the cow's starting point. A and B are the feet of the perpendiculars from C and G to ~, and H is the point on GB such that B is the midpoint of GH. The shortest distance between two points is along the straight line joining them, so the cow's route should be along a straight line from C to a point P on ~ and then along PG. The line segments PG and PH are equal in length; if P does not equal B, this follows from the fact that triangles PBH and PBG are congruent. Let D be the point where CH meets ~.   Since CH is a straight line, CD + DH > CP + PH for any point P (other than D) on ~. Hence CD + DG < CP + PG and so the cow should walk straight from C to D and then straight from D to G.

(Since triangles ADC and BDH are similar, AD/BD = 1/2 and so D is 10m from A.)
 
 

 

 
 
 
 

 S3. Let n be a positive integer. To find all positive integers x, y satisfying the equation xy = n(x + y) (1)

write X = x - n and Y = y - n. Then (1) is equivalent to (X + n)(Y + n) = n(X + Y + 2n) and so to XY = n2  The solutions (in positive integers) of this are X = m, Y =  , where m is any factor of n2 .

When n = 1999, n2 = 19992 , which has factors 1, 1999 and 19992 (1999 is a prime number). Hence the solutions of XY = 19992 are (i) X = 1, Y = 19992 ; (ii) X = 1999 , Y = 1999; and (iii) X = 19992, Y = 1 . Since X = x - 1999 and Y = y - 1999, and 1999   2000 = 3,998,000,

the solutions of xy = 1999(x + y) in positive integers are:
 
 

(i) x = 2,000, y = 3,998,000;

(ii) x = 3998, y = 3998;

(iii) x = 3,998,000, y = 2000 .
 
 

 Alternative method for S3.

Suppose that xy = 1999(x + y). Since 1999 is a prime number, either x or y is a multiple of 1999.

Case 1 Suppose that x = 1999u, where u is a positive integer. Then 1999uy = 1999(1999u + y). Therefore uy = 1999u + y and so y =  . Hence, since 1999 is a prime, either u - 1 = 1999 or u - 1 = 1. In the first case, u = 2000, which leads to the solution x = 1999   2000, and in the second u = 2, which leads to the solution x = 1999   2, y = 1999   2.

Case 2 Suppose now that y is a multiple of 1999. Then, by a similar argument, we obtain the solutions x = 2000, y = 1999 2000 and x = 1999   2, y = 1999   2.

The second of these is one of the solutions already obtained.
 
 

S4 
 
 

Extend the line AB to F such that BF = BC. Triangle DBF and triangle DBC have two pairs of equal corresponding sides. The included angles are FBD and CBD; and angle FBD

= 180 - ABD (straight line at B)

= 180 - ACD (angles on arc AD are equal)

= 180 - CAD (triangle ADC is isosceles)

= CBD (opposite angles of a cyclic quadrilateral are supplementary).

Hence FBD = CBD. It follows that triangle DBF is congruent to triangle DBC and thus DC = DF. But triangle ADC is isosceles, and so DC = DA, whence DF = DA and therefore triangle ADF is isosceles, which shows that E is the midpoint of AF. Hence AE = EF = EB + BC and therefore E is the midpoint of the path from A to C along the line segments AB and BC as required.

Alternative proof for S4, using trigonometry

Let angle BAD = a and angle DBA = b . Then AE =  and BE =  . The points A, C, B and D are concyclic, and so angle DCA = angle DBA = b . Since D is the mid-point of the arc CA, D ADC is isosceles and therefore angle DAC = b . Hence angle BAC = b  - a . Since A, C, B and D are concyclic, angle BDC = angle BAC = b -a .
Applying the sine rule to triangle BCD we get BC = BD ´ .
It follows that

AE - BE - BC  =  DE

=  DE

=  0 since sin(b - a ) = sinb cosa  - cosb sina .

Therefore AE = BE + BC .