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Individual Contest
Time limit: 120 minutes 2015/12/11 Section A.
In this section, there are 12 questions. Fill in the correct answer on the space provided at the end of each question. Each correct answer is worth 5 points. Be sure to read carefully exactly what the question is asking.
1. Evaluate 1 1 1 1 9 80 80 79 79 78 10 3 M = − + − − − − − " − .【Submitted by Bulgaria_FPMG】 【Solution】 Since 1 1 1 n n
n− n− = + − for all n, the expression is equal to
9+ 80 − 80− 79 + 79+ 78 ...− − 10− =3 6.
ANS: 6
2. Find the smallest positive integer n such that both 2n and 3n+ are squares of 1 integers.【Submitted by VIETNAM】
【Solution】
Let n be such that 3n+ = and 1 x2 n=2y2 for some positive integers x and y. Then
2 2
6 1
x = y + . It follows that x is odd and y is even. The smallest possible value of y is
2, and the corresponding value of x is 5. Hence the smallest possible value of n is 8.
ANS: 8
3. How many different possible values of the integer a are there so that
||x− − −2 | | 3 x||= −2 a has solutions?【Submitted by Bulgaria_SMG】
【Solution】
The expression ||x− − −2 | | 3 x|| measures the difference between the distance from
x to 2 and the distance from x to 3. If x is not between 2 and 3, the difference is
always 1, so that we may have a=1. If x is between 2 and 3, the difference is less than 1, and takes only the integer value 0 at x=2.5. Hence we may also have a=2. These are the only 2 possible values.
ANS: 2
4. If k −9 and k+36 are both positive integers, what is the sum of all possible values of k?【Submitted by VARNA】
【Solution】
Let n2 = +k 36 and m2 = − then k 9 n2−m2 =45 and (n−m n)( +m)=45. The divisors of 45 are 1, 3, 5, 9, 15 and 45. There are three options. If n+ =m 45 and
1
n− = then m k =493. If n+ = and m 15 n− = then m 3 k =45. If n+ = m 9 and 5n− = then m k =13. The desired sum is 493 45 13+ + =551.
5. Find the largest positive integer n such that the sum of the squares of the positive divisors of n is n2 +2n+ .【Submitted by Bulgaria_SMG】 2
【Solution】
If n is prime, then the sum of squares of its divisors is n2+ , which is too small. 1 If n is the square of a prime, then the sum is n2 + + , 1 less than the desired value. n 1 If n is a higher power of a prime, the sum will be too large. Hence n=ab where a and b are relatively prime integers greater than 1. Then the sum of the squares of its positive divisors is at least n2 +a2 +b2 + ≥1 n2 +2ab+ =1 n2 +2n+ . Thus a and b 1 must be prime numbers so that they are the only other positive divisors. Moreover, we must have 1=a2 +b2 −2ab=(a−b)2 so that a− = . Thus the only possibility b 1 is (a, b) = (3, 2) , so that n= is the largest, and in fact, the only solution. 6
ANS: 6
6. Find the smallest two-digit number such that its cube ends with the digits of the original number in reverse order.【Submitted by MALAYSIA】
【Solution】
The cube of the number 10a+ is b (10a+b)3 =1000a3 +300a b2 +30ab2 + . Then b3
the units digit of b is a. If 3 b= , 2, 3, 4, 5, 6, 7, 8 or 9, then 1 a = , 8, 7, 4, 5, 6, 3, 2 1 or 9 and 30ab2 +b3 =31, 968, 1917, 1984, 3875, 6696, 4753, 4352 or 22599
respectively. Thus the only possibility is 99.
ANS: 99
7. A Mathematics test consists of 3 problems, each problem being graded independently with integer points from 0 to 10. Find the number of ways in which the total number of points for this test is exactly 21.【Submitted by ROMANIA】
【Solution】
If the first problem is graded 10, then there are 10 ways to grade the other two problems in order to obtain 21 points 1 10+ = + = + = + = + = + = 2 9 3 8 4 7 5 6 6 5
7+ = + = + =4 8 3 9 2 10 1 21+ = . If the first problem is graded 9, then there are 9 ways to grade the other two problems in order to obtain 21 points, and so on. If the first problem is graded 1, then there is 1 way to grade the other two problems in order to obtain 21 points. The total number of ways is thus: 10+ + +9 8 "+ =1 55.
ANS: 55 ways
8. In the triangle ABC, the bisectors of ∠CAB and ∠ABC meet at the in-center I. The extension of AI meets the circumcircle of triangle ABC at D. Let P be the foot of the perpendicular from B onto AD, and Q a point on the extension of AD such that ID=DQ. Determine the value of BQ IB
BP ID × × .【2013 AITMO Proposal】 Q P A B C D I
【Solution】
Let us prove that △BPQ is similar to △IBQ. First ∠IAB= ∠CAD= ∠CBD.
As ∠IBA= ∠IBC, we have ∠IAB+ ∠IBA= ∠CBD+ ∠IBC so that
DIB DBI
∠ = ∠ , thus DI =DB=DQ. This means △IBQ is a right triangle with
90
IBQ
∠ = °. As Q∠ is a common angle, we thus have △BPQ is similar to △IBQ.
Hence, BQ IQ 2ID BP = IB = IB . Consequently, 2 BQ IB BP ID × = × . ANS: 2
9. D and E are points inside an equilateral triangle ABC such that D is closer to AB than to AC. If AD=DB=AE =EC = cm and 7 DE = cm, what is the length of 2 BC, in cm? 【Submitted by Philippines】
【Solution】
Let H and K be the respective feet of perpendicular from A and D. Let the side length of ABC be x. Then
2 x BH = , 1 2 x BK = − and 72 ( 1)2 2 x DK = − − . Furthermore, we have 3 2 x AH = and 4 3 AH −DK = . From 49 ( 1)2 3 4 3 2 2 x x − − = − , we have 2 2 3 12 4 4 x x x x − + = − , which yields x= . 13 ANS: 13 cm
10. In a class, five students are on duty every day. Over a period of 30 school days, every two students will be on duty together on exactly one day. How many students are in the class?【Submitted by ROMANIA】
【Solution】
Let n be the number of students in the class and let N be the number of pairs of students that are on duty on each of the 30 days. Since there were 5 students on duty every day, 5 30 300
2
N =⎛ ⎞⎜ ⎟× =
⎝ ⎠ . On the other hand, since two students are on duty on
exactly 1 days, ( 1) 2 2 n n n N =⎛ ⎞⎜ ⎟= × − ⎝ ⎠ . From 2 0=n − −n 600=(n+24)(n−25), we have n=25. H K C B A D E
Arrange the students in a 5×5 array as follows. A B C D E F G H I J K M N O P Q R S T U V W X Y Z
We divide the 30 days into 6 blocks of 5 days each, with the following duty schedule. Block 1 ABCDE FGHIJ KMNOP QRSTU VWXYZ
Block 2 AGNTZ BHOUV CIPQW DJKRX EFMSY Block 3 AHPRY BIKSZ CJMTV DFNUW EGOQX Block 4 AIMUX BJNQY CFORZ DGPSV EHKTW Block 5 AJOSW BFPTX CGKUY DHMQZ EINRV Block 6 AFKQV BGMRW CHNTX DIOTY EJPUZ
This structure is called a (30,25,6,5,1) block design, and is based on the finite affine geometry of order 5.
ANS: 25 students
11. A committee is to be chosen from 4 girls and 5 boys and it must contain at least 2 girls. How many different committees can be formed? 【Submitted by
ROMANIA】
【Solution】
There is 1 way to include all 4 girls, 4 ways to include 3 of them and 6 ways to include 2 of them. Thus the girls can be chosen in 1 4+ + = ways. The boy can 6 11 be chosen in 25 =32 ways. Hence a total of 11 32× =352 different committee can be formed.
ANS: 352 committee
12. Find the largest positive integer such that none of its digits is 0, the sum of its digits is 16 but the sum of the digits of the number twice as large is less than 20.
【Submitted by Russia】
【Solution】
In adding the number to itself, there must be carries over as otherwise the sum of the digits of the number twice as large will be 32. A carry over reduces the sum of the digits by 9, so we need at least two of them. The largest such number is 55111111.
Section B.
Answer the following 3 questions. Show your detailed solution on the space provided after each question. Each question is worth 20 points.
1. What is the number of ordered pairs (x, y) of positive integers such that 3 1 1 2
x + = y
and xy ≥3 6?【Submitted by Bulgaria_SMG】 【Solution】
We have 3 1 1 6 2 ( 6)( 2) 12 2 y x xy x y
x + = ⇔y + = ⇔ − − = . (5 points)
So (x, y) = (7, 14), (8, 8), (9, 6), (10, 5), (12, 4) or (18, 3). (5 points)
The respective values of xy are 98, 64, 54, 50, 48 and 54. (5 points)
Since (3 6)2 =54, we must reject (10, 5) and (12, 4), leaving behind 4 ordered pairs.
(5 points)
Ans: 4
2. What is the minimum number of the 900 three-digit numbers we must draw at random such that there are always seven of them with the same digit-sum?
【Submitted by ROMANIA】
【Solution】
The minimum digit-sum is 1 and the maximum digit sum is 27. Each of these appears once, in 100 and 999 respectively. Each of 2 and 26 appears three times, in 101, 110 and 200 in the former case and in 998, 989 and 899 in the latter case(5 points). Similarly, each of 3 and 25 appears six times. (5 points) Every other digit-sum appears at least seven times. We may draw 2 1 2 3× + × +23 6 146× = numbers without having seven with the same digit-sum. However, if we draw another number, we are sure to have seven numbers with the same digit-sum. (10 points)
ANS: 147
3. Point M is the midpoint of the semicircle of diameter AC. Point N is the midpoint of the semicircle of diameter BC and P is midpoint of AB.
Prove that ∠PMN = ° .45 【Submitted by VARNA】
C
B A P
M
【Solution 1】
Let points D and E be the midpoints of AC and BC respectively. Then
DM =DC=PE and DP=CE =NE.(5 points)Moreover, 90∠MDP = ° + ∠ADP
while ∠PEN = ° + ∠90 PEB. Hence triangles MPD and PNE are congruent. (5 points) We have MP=PN and PMD∠ = ∠EPN. It follows that
180 90 90
MPN MPD DPE EPN MPD ADP PMD
∠ = ∠ + ∠ + ∠ = ∠ + ∠ + ∠ = ° − ° = °,
(5 points)so that ∠PMN = ∠PNM =45°.(5 points)
【Solution 2】
Construct the perpendicular line MD and NE, where D is the midpoint of AC and E is the midpoint of BC. Since DM =DC, we have ∠DMC= ∠DCM = ° . Similarly, 45
45
ECN
∠ = °.(5 points)
D and P are the midpoints of AC and AB, respectively, so DP BC and //
1 2
DP= BC=CE. Since DP BC , // ∠ADP= ∠ACB. (5 points) Thus we have 90 45 45
MDP ADP ACB DCM ECN ACB MCN
∠ = ° + ∠ = ° + ° + ∠ = ∠ + ∠ + ∠ = ∠ . And since CM 2 CE DP
DM = =CN =CN , triangle MDP and MCN are similar triangles. (5
points) Hecnce ∠PMD= ∠NMC. So 45∠PMN = ∠PMD+ ∠DMN = ∠NMC+ ∠DMN = ∠DMC= °(5 points) E D C B A P M N E D C B A P M N
