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Stage 7 · Algebra

10 problems · National olympiad second round; IMO P1/P4 · mathsolympiadprep.com

The answer key prints on its own page at the end.

  1. P(x)=ax2+bx+cP(x)=ax^2+bx+c has exactly 11 different real root where a,b,ca,b,c are real numbers. If P(P(P(x)))P(P(P(x))) has exactly 33 different real roots, what is the minimum possible value of abcabc?

    1. A-3
    2. B-2
    3. C232\sqrt 3
    4. D333\sqrt 3
    5. ENone of above\text{None of above}

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  2. Solve the following system of equations for real x,yx,y and zz:
    \text{}
    x &=& 2y+3\sqrt{2y+3}\\
    y &=& 2z+3\sqrt{2z+3}\\
    z &=& 2x+3 .\text{2x+3 .}

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  3. Determine all functions f:ZZf: \mathbb{Z} \rightarrow \mathbb{Z} satisfying f(f(m)+n)+f(m)=f(n)+f(3m)+2014 f(f(m)+n)+f(m)=f(n)+f(3 m)+2014 for all integers mm and nn. (Netherlands) Answer. There is only one such function, namely n2n+1007n \longmapsto 2 n+1007.

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  4. Find the smallest real number CC, such that for any positive integers xyx \neq y holds the following:

    min({x2+2y},{y2+2x})<C\min(\{\sqrt{x^2 + 2y}\}, \{\sqrt{y^2 + 2x}\})<C

    Here {x}\{x\} denotes the fractional part of xx. For example, {3.14}=0.14\{3.14\} = 0.14.

    Proposed by Anton Trygub

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  5. A polynomial p(x)p(x) with real coefficients is said to be almeriense if it is of the form:

    p(x)=x3+ax2+bx+a p(x) = x^3+ax^2+bx+a

    And its three roots are positive real numbers in arithmetic progression. Find all almeriense polynomials such that p(74)=0p\left(\frac{7}{4}\right) = 0

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  6. a,b,ca, b, c are positive real numbers such that (ab1)(bc1)(ca1)=1(\sqrt {ab}-1)(\sqrt {bc}-1)(\sqrt {ca}-1)=1
    At most, how many of the numbers: abc,acb,bac,bca,cab,cbaa-\frac {b}{c}, a-\frac {c}{b}, b-\frac {a}{c}, b-\frac {c}{a}, c-\frac {a}{b}, c-\frac {b}{a} can be bigger than 11?

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  7. p1. Find all real numbers that satisfy the equation (1+x2+x4+....+x2014)(x2016+1)=2016x2015(1 + x^2 + x^4 + .... + x^{2014})(x^{2016} + 1) = 2016x^{2015}

    p2. Let AA be an integer and A=2+20+201+2016+20162+...+20162016...201640digitsA = 2 + 20 + 201 + 2016 + 20162 + ... + \underbrace{20162016...2016}_{40\,\, digits}
    Find the last seven digits of AA, in order from millions to units.

    p3. In triangle ABCABC, points PP and QQ are on sides of BCBC so that the length of BPBP is equal to CQCQ, BAP=CAQ\angle BAP = \angle CAQ and APB\angle APB is acute. Is triangle ABCABC isosceles? Write down your reasons.

    p4. Ayu is about to open the suitcase but she forgets the key. The suitcase code consists of nine digits, namely four 00s (zero) and five 11s. Ayu remembers that no four consecutive numbers are the same. How many codes might have to try to make sure the suitcase is open?

    p5. Fulan keeps 100100 turkeys with the weight of the ii-th turkey, being xix_i for i{1,2,3,...,100}i\in\{1, 2, 3, ... , 100\}. The weight of the ii-th turkey in grams is assumed to follow the function xi(t)=Sit+200ix_i(t) = S_it + 200 - i where tt represents the time in days and SiS_i is the ii-th term of an arithmetic sequence where the first term is a positive number aa with a difference of b=15b =\frac15. It is known that the average data on the weight of the hundred turkeys at t=at = a is 150.5150.5 grams. Calculate the median weight of the turkey at time t=20t = 20 days.

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  8. 165 Find the integer solution of the equation x+x+x++x1964 terms =y\underbrace{\sqrt{x+\sqrt{x+\sqrt{x+\cdots+\sqrt{x}}}}}_{1964 \text { terms }}=y.

    untranslated part:
    将上面的文本翻译成英文,请保留源文本的换行和格式,直接输出翻译结果。

    translated part:
    Find the integer solution of the equation x+x+x++x1964 terms =y\underbrace{\sqrt{x+\sqrt{x+\sqrt{x+\cdots+\sqrt{x}}}}}_{1964 \text { terms }}=y.

    4. 165

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  9. Let mm and nn two given integers. Ana thinks of a pair of real numbers xx, yy and then she tells Beto the values of xm+ymx^m+y^m and xn+ynx^n+y^n, in this order. Beto's goal is to determine the value of xyxy using that information. Find all values of mm and nn for which it is possible for Beto to fulfill his wish, whatever numbers that Ana had chosen.

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  10. The function f(n)f(n) is defined on the positive integers and takes non-negative integer values. f(2)=0,f(3)>0,f(9999)=3333f(2)=0,f(3)>0,f(9999)=3333 and for all m,n:m,n: f(m+n)f(m)f(n)=0 or 1. f(m+n)-f(m)-f(n)=0 \text{ or } 1. Determine f(1982)f(1982).

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Answer key — Stage 7 · Algebra

Worked solutions for every problem are on the site, one page per problem.

  1. 2-2 open
  2. x=y=z=3x = y = z = 3 open
  3. f(n)=2n+1007f(n) = 2n + 1007 open
  4. φ1\varphi - 1 open
  5. p(x)=x3214x2+738x214p(x) = x^3 - \frac{21}{4}x^2 + \frac{73}{8}x - \frac{21}{4} open
  6. 44 open
  7. 200.5200.5 open
  8. x=0,y=0x=0, y=0 open
  9. (m,n)=(2k+1,2t(2k+1))(m, n) = (2k+1, 2t(2k+1)) open
  10. 660660 open

Problems belong to the competitions that set them and are reproduced from open datasets under their licences; every problem page names its source. Free to copy for classroom use.