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HANOI OPEN MATHEMATICAL OLYMPIAD 2017 - Junior level

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#1
IMO20xx

IMO20xx

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HANOI OPEN MATHEMATICAL OLYMPIAD 2017

Junior level

  1. Suppose $x_1,x_2,x_3$ are the roots of polynomial $$P(x)=x^3-6x^2+5x+12.$$ Then $|x_1|+|x_2|+|x_3|$ is

A.    $4$           

B.     $6$

C.     $8$

D.    $14$

E.     None of the above

  1. How many pairs of positive integers $(x,y)$ are there, those satisfy the identity $$2^x-y^2=1?$$

A.    $1$

B.     $2$

C.     $3$

D.    $4$

E.     None of the above

  1. Suppose $n^2+4n+25$ is a perfect square. How many such non-negative integers $n$’s are there?

A.    $1$

B.     $2$

C.     $4$

D.    $6$

E.     None of the above

  1. Put $$S=2^1+3^5+4^9+5^{13}+\cdots +505^{2013}+506^{2017}.$$ The last digit of $S$ is

A.    $1$

B.     $3$

C.     $5$

D.    $7$

E.     None of the above

  1. Let $a,b,c$ be two-digit, three-digit, four-digit numbers respectively. Assume that the sum of all digits of numbers $a+b$, and the sum of all digits of number $b+c$ are equal to $2$. The largest value of $a+b+c$ is

A.    $1099$

B.     $2099$

C.     $1199$

D.    $2199$

E.     None of the above

  1. Find all triples of positive integers $(m,n,p)$ such that $$2^mp^2+27=q^3,$$ and $p$ is a prime.
  2. Determine the two last digits of number $$Q=2^{2017}+2017^2.$$
  3. Determine all real solutions $x,y,z$ of the following system of equations $$\begin{cases} x^3-3x &=4-y \\ 2y^3-6y &=6-z \\ 3z^3-9z &=8-x. \end{cases}$$
  4. Prove that every equilateral triangle of area $1$ can be covered by $5$ arbitrary equilateral triangles which have the total area of $2$.
  5. Find all non-negative integers $a,b,c$ such that the roots of equations $$x^2-2ax+b=0,\quad (1)$$ $$x^2-2bx+c=0,\quad (2)$$ $$x^2-2cx+a=0.\quad (3)$$ are non-negative integers.
  6. Let $S$ denote a square of the side-length $7$ and let $8$ squares of the side-length $3$ be given. Show that $S$ can be covered by those $8$ small squares.
  7. Does there exist a sequence of $2017$ consecutive integers which contains exactly $17$ primes?
  8. Let $a,b,c$ be the side-lengths of triangle $ABC$ which $a+b+c=12$. Determine the smallest value of $$M=\frac{a}{b+c-a}+\frac{4b}{c+a-b}+\frac{9a}{a+b-c}.$$
  9. Given trapezoid $ABCD$ which bases $AB\parallel CD$ ($AB<CD$). Let $O$ be the intersection of $AC$ and $BD$. Two straight lines from $D$ and $C$ are perpendicular to $AC$ and $BD$ intersect at $E$, i.e. $CE\perp BD$ and $DE\perp AC$. By analogy, $AF\perp BD$ and $BF\perp AC$. Are three points $E,O,F$ collinear?
  10. Show that an arbitrary quadrilateral can be devided into $9$ isosceles triangles.

 

END OF PAPER



#2
Olympusreacher

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Three first answer

1.P(x)=x3-6x2+5x+12=(x+1)(x-3)(x-4)=0

 

So the roots of the polynomial are x1=-1,x2=4,x3=3

The sum of their abosolute value is 8

So (D) is true

2. Solve the diophantine equation:2x-y2=1

⇔2x=1-y2>0

⇔y2<1

 -1<y<1

⇔y=0

⇔2x= 1

⇔x=0

So there is only one pair (x,y) is (0,0)

Answer is (A)

3.Solve the equation

Call the perfect square is y

We have the equation:$y^2=n^2+4n+25 ⇔y^2=(x+2)^2+21$

We have 4 pairs (x,y)=(8,11),(-12,-11),(0,5),(-4,-5)

So there are 2 number of x is non-negative intergers

(B) is the answer.

 


Weak people revenge, strong people forgive, intelligent people ignore.

                                                                                                                                                                   Albert Einstein


#3
viethoang2002

viethoang2002

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Three first answer

1.P(x)=x3-6x2+5x+12=(x+1)(x-3)(x-4)=0

 

So the roots of the polynomial are x1=-1,x2=4,x3=3

The sum of their abosolute value is 8

So (D) is true

 

câu đó bạn ấn nhầm là (d) chứ (c) mới đúng :v

câu 4 và 5 đều có đáp án là (e) nhá ^^


Bài viết đã được chỉnh sửa nội dung bởi viethoang2002: 08-07-2017 - 12:54


#4
viethoang2002

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2. Solve the diophantine equation:2x-y2=1

⇔2x=1-y2>0

⇔y2<1

 -1<y<1

⇔y=0

⇔2x= 1

⇔x=0

So there is only one pair (x,y) is (0,0)

Answer is (A)

 

câu này bạn sai dấu ở khúc chuyển vế :v làm như sau

biến đổi thành (y-1)(y+1)=$2(2^{x-1}-1)$

suy ra y lẻ suy ra vế trái chia hết cho 4 do đó vế phải chia hết cho 4 (vô lý )

từ đó ta chọn a 


Bài viết đã được chỉnh sửa nội dung bởi viethoang2002: 08-07-2017 - 13:55





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