Skip to main content

2006/07 AHSMC Part 1

All 16 problems. Take it as a timed test — 1:20:00 on the clock, scored the way the AMC is — or read through the problems and open any of them for its worked solution.

16 problems · 6 points correct, 1.5 for blank, 0 for wrong

  1. The value of 24488162^4 4^8 8^{16} is

  2. The number of non-congruent rectangles with integer sides and area 2006=2×17×592006 = 2 \times 17 \times 59 is

  3. The number of pairs (m,n)(m, n) of positive integers such that m4+n=100000001m^4 + n = 100000001 is

  4. In a city, all streets run north-south or east-west, dividing the city into squares. AA, BB, CC and DD are four students who live at four street intersections which define a rectangle, with AA and CC at opposite corners of this rectangle. They all go to the same school, which is at some street intersection within this rectangle. Each goes to school by a most direct route along streets. AA travels 1010 blocks, BB 2020 blocks and CC 5050 blocks. The number of blocks DD travels is

  5. When (1+x)(1+x2)(1+x3)(1+x4)(1+x5)(1 + x)(1 + x^2)(1 + x^3)(1 + x^4)(1 + x^5) is expanded, the coefficient of the term x9x^9 is

  6. On a plane are two points AA and BB at a distance 55 apart. The number of straight lines in this plane which are at a distance 22 from AA and at a distance 33 from BB is

  7. A country has three provinces, each province has three cities, each city has three wards and each ward has three electors. In a two-way election, a candidate wins a ward by getting more votes in the ward, wins a city by winning more wards in the city, wins a province by winning more cities in the province, and wins the election by winning more provinces. Only electors may vote, and they must vote. The minimum number of votes needed to guarantee winning the election is

  8. The teacher asked, “What is the largest possible diameter of a circular coin of negligible thickness which may be stored in a rectangular box with inner dimensions 7×8×97 \times 8 \times 9?” Ace said less than 88, Bea said 88, Cec said strictly between 88 and 99, Dee said 99 and Eve said more than 99. The one who was right was

  9. If aa and bb are two real numbers which satisfy a+b−ab=1a + b - ab = 1 and aa is not an integer, then bb

  10. The non-zero numbers aa, bb, cc, xx, yy and zz are such that xa=yb=zc\dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c}. The value of xyz(a+b)(b+c)(c+a)abc(x+y)(y+z)(z+x) \dfrac{xyz(a + b)(b + c)(c + a)}{abc(x + y)(y + z)(z + x)} is

  11. If kk is a rational number such that 93−1123=3+k2\sqrt[3]{9\sqrt{3} - 11\sqrt{2}} = \sqrt{3} + k\sqrt{2}, then the value of kk is

  12. All positive integers which can be expressed as a sum of one or more different integer powers of 55 are written in increasing order. The first three terms of this sequence are 11, 55 and 66. The fiftieth term is

  13. Consider all polynomials whose coefficients are all integers, whose roots include 32\dfrac{\sqrt{3}}{2} and 23\dfrac{\sqrt{2}}{3} and whose degree is as small as possible. Among the coefficients of these polynomials, the smallest positive coefficient is

  14. Colleen used a calculator to compute a+bc\dfrac{a + b}{c}, where aa, bb and cc are positive integers. She pressed aa, ++, bb, //, cc and == in that order, and got the answer 1111. When she pressed bb, ++, aa, //, cc and == in that order, she was surprised to get a different answer 1414. Then she realized that the calculator performed the division before the addition. So she pressed ((, aa, ++, bb, )), //, cc and == in that order. She finally got the correct answer, which is

  15. The base of a tetrahedron is an equilateral triangle of side 11. The fourth vertex is at a distance 11 above the centre of the base. The radius of the sphere which passes through all four vertices of the tetrahedron is

  16. If p(x)p(x) is a polynomial of degree 44 such that p(−1)=p(1)=5p(-1) = p(1) = 5 and p(−2)=p(0)=p(2)=2p(-2) = p(0) = p(2) = 2, then the maximum value of p(x)p(x) is

0 of 16 answered. Each problem left blank still scores 1.5 points.