Tournament of Towns, 1983-1984, Fall, Practice Version, Grades 9-10

• Question 1 - Cherries and Blueberries

  `175` kg of cherries cost more than `125` kg of blueberries, but less than `126` kg of blueberries. In addition, it is known that a kilogram of cherries costs a whole number of shekels, and a kilogram of blueberries also costs a whole number of shekels.

  Prove that `80` shekels is not enough to buy one kilogram of blueberries and three kilograms of cherries.

  S. Fomin

  Topics:

  Number Theory Arithmetic Algebra -> Inequalities Algebra -> Word Problems

• Question 2 - Pentagon

  In a convex pentagon `ABCDE`, the following holds: `AE=AD`, `AB=AC`, and `angle CAD=angle ABE + angle AEB`.

  Let `AM` be the median to the side `BE` in the triangle `ABE`. Prove that `AM` is half the length of the segment `CD`.

  Topics:

  Geometry -> Plane Geometry -> Angle Calculation Geometry -> Plane Geometry -> Triangles -> Triangle Congruence Geometry -> Plane Geometry -> Plane Transformations -> Congruence Transformations (Isometries) -> Rotation

• Question 3 - Frame

  On a grid paper, a square of size `NxxN` is given. Consider its frame with a width of one square. It consists of `4*(N-1)` squares.

  Can you write `4*(N-1)` consecutive integers (not necessarily positive) in the squares of the frame, such that the following condition holds:

  For every rectangle whose vertices are on the frame and whose sides are parallel to the diagonals of the original square, the sum of the numbers at the vertices is equal to a constant value. This also includes the "degenerate" rectangles of zero width that coincide with the diagonals of the square - in this case, simply sum the two numbers at the opposite vertices of the square.

  For:

  a. `N=3`

  b. `N=4`

  c. `N=5`

  

  Topics:

  Arithmetic Number Theory -> Division -> Parity (Even/Odd) Proof and Example -> Constructing an Example / Counterexample Algebra -> Sequences -> Arithmetic Progression / Arithmetic Sequence Combinatorics -> Case Analysis / Checking Cases -> Processes / Procedures

• Question 4 - Product of Digits

  For a natural number `x`, let `P(x)` denote the product of the digits of `x`, and let `S(x)` denote the sum of the digits of `x`. 

  How many solutions are there to the equation:

  `P(P(x))+P(S(x))+S(P(x))+S(S(x))=1984`

  Topics:

  Number Theory Algebra -> Equations

• Question 4

  Move one matchstick to make a valid equation.

  

  Note: There may be multiple solutions.

  Topics:

  Arithmetic Puzzles and Rebuses -> Matchstick Puzzles