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Team Contest
1. A function f x y( , ) satisfies f x y( , )+ f x z( , )= f x y( , + +z 3) and
( , ) ( , ) ( 4, )
f x y + f z y = f x+ +z y . Find the value of f(2016, 2017) if (1, 1) 1 f = . Solution We have (2016, ) (1, ) (2011, ) 2 (1, ) (2006, ) 403 (1, ) (1, ) 404 (1, ) f y f y f y f y f y f y f y f y = + = × + = × + = × ⋮
So f(2016, 2017)=404× f(1, 2017). And we also have (1, 2017) (1, 1) (1, 2013) 2 (1, 1) (1, 2009) 504 (1, 1) (1, 1) 505 (1, 1) f f f f f f f f = + = × + = × + = × ⋮ Hence f(2016, 2017)=404 505× × f(1, 1)=204020. ANS: 204020
2. In a rectangular coordinate plane, straight line L passes through the point (14, 15), the
x-coordinate in the intersection point of L and x-axis is a prime number, the y-coordinate in the intersection point of L and y-axis is a non-zero positive integer. How many possible straight line L will meet all the given
conditions? 【Submitted by Philippines】 Solution
Let the intersection point of L and x-axis denoted as (p, 0), the intersection point of L and y-axis
denoted as (0, q), where p is a prime number greater than 14 and q is a nonzero positive integer greater than 15.
International Young Mathematicians’ Convention
Junior Level
(14, 15) (0, q) (p, 0) x L ySince (p, 0), (14, 15), (0, q) must be collinear, then the slope of L is 0 15 15 14 14 0 q p − = − − − . So that (p−14)(15−q)= −( 15) 14× = −210, which is equivalent as
(p−14)(q−15) =210. Observe that p−14 is an odd number since p is an odd
prime number. It follows that 14 p− 1 3 5 7 15 21 35 105 15 q− 210 70 42 30 14 10 6 2 or p 15 17 19 21 29 35 49 119 q 225 85 57 45 29 25 21 17
Hence, (p, q) = (17, 85), (19, 57) or (29, 29). There are 3 possible solutions.
ANS: 3 【Marking Scheme】
Describe the slope of L, 10 marks.
Find the relation (p−14)(q−15)=210, 10 marks.
List all correct integral pairs (p, q), 10 marks.
Note that if the student didn’t observe that p−14 is an odd number, then he/she
has to list 16 pairs.
3 marks for each possible solution.
Get all of the possible solutions without wrong solutions, 10 marks.
3. A, B, C, D and E participated in a marathon. At the start, A is leading, B is in the second place, C is in the middle, D is in the fourth place and E is last. During the marathon, A trades place with the other four runners a total of 11 times, while B does that 11 times, C does that 22 times and E does that 44 times. If A finished ahead of D, who is in the second place among these five runners at the finish? Solution:
When a runner trade places in a race, an odd numbered position becomes even numbered, and vice versa.
Since A starts in an odd-numbered place and trades places an odd number of times, he must finish in an even-numbered position.
Since B starts in an even-numbered place and trades places an odd number of times, he must finish in an odd-numbered position.
Since C starts in an odd-numbered place and trades places an even number of times, he must finish in an odd-numbered position.
Since E starts in an odd-numbered place and trades places an even number of times, he must finish in an odd-numbered position.
So D must finish in an even-numbered position. Since A finished ahead of D, A is in the second place and D is in the fourth place at the finish.
4. In the acute triangle ABC, ∠ = °A 60 . Two altitudes, BE and CF, intersect at point H. Point O is the
circucentre of triangle ABC, as shown in the diagram below. Find the measure of ∠FHO, in degrees. Solution
Connect OB and OC. Observe that
2 120
BOC BAC
∠ = ∠ = °.
Also ∠BHC = ∠FHE =180° − ∠BAC =180° − ° =60 120°.
Hence BCHO is cyclic and ∠OHB = ∠OCB = °30 since OB =OC.
On the other hand, ∠BHF =180° − ∠BHC =180° −120° = °60 . It follows that
60 30 30
FHO
∠ = ° − ° = °.
Answer: 30° 【Marking Scheme】
Observe that ∠BOC =120°, 10 marks.
Observe that ∠BHC =120°, 10 marks.
Observe that ∠OHB = ∠OCB= °30 , 10 marks.
Conclude that ∠BHF = °60 , 5 marks.
Conclude that ∠FHO= °30 , 5 marks.
5. A code of the coffer is an arrangement of numbers satisfied that the differences between neighbours are all different. For example, the numbers
have differences 3, 2 and 1 ― all different. Now the code is the numbers from 1 to 6 that are arranged with the condition, and with 4 in the fourth position from left to right,
Find the sum of the possible values of the rightmost position of the code. Solution
When placing the numbers 1, 2, 3, 4, 5 and 6 in six boxes, there are five differences, which we will write between the boxes:
? ? ? ? ? ? 4 ? ? ? ?
Since the only possible differences are 1, 2, 3, 4 and 5, each must appear once. The difference of 5 must come from the pair
1 5 6 (possibly reversed) and the difference of 4 from either
1 4 5 or 2 4 6
These always have one number in common, so they must be in the first three boxes. That is, one of
A C H O B E F 1 4 2 3 4
The difference of 3 must come from one of
1 3 4 、 2 3 5 or 3 3 6
Observe that the last two cases can’t be in the last two boxes since 6 and one of 2 and 5 must be in the first three boxes. So, the only way to fit one of these with one of the four possibilities for the first three boxes is
2 4 6 5 1 3 4 ? ? ? ?
The final two numbers can be placed either way around, giving two codes: 2 4 6 5 1 3 4 2 3 1 5 and 2 4 6 5 1 3 4 1 5 2 3 So the sum of the possible values of the rightmost position of the code is 5 3 8+ = .
Answer: 8 6. A positive integer consists of distinct digits each of which is a divisor of the
number. What is the largest number that fit the condition? Solution
Clearly, none of the digits may be 0. If one of them is 5, then the units digit of the number must be 5, and the number will not be divisible by 2, 4, 6 or 8. This leaves a maximum of 5 different digits, which we shall see is short of the maximum. Thus it is better to leave out 5. Note that 1 + 2 + 3 + 4 + 6 + 7 + 8 + 9 = 40, which is not a multiple of 3. The first multiple of 9 below 40 is 36. By leaving out 4, we may have as many as 7 different digits.
Now we want to find the largest 7-digit number so we can assume the last three digits are formed by 1, 2 and 3 first. Since the 7-digit number is a multiple of 8, 312 is the end of the 7-digit. Since 9876312 is not a multiple of 7 and 9867312 = 7 × 1409616, the largest number that fit the condition is 9867312.
Answer: 9867312 【Marking Scheme】
Observe that none of the digits may be 0, 5 marks. Observe that it is better to leave out 5, 5 marks. Observe that it is better to leave out 4, 10 marks. Observe that 312 is the end of the 7-digit, 10 marks. Conclude that the answer is 9867312, 10 marks.
