Math Olympiad Problems for Grades 5 and 6: Problems with Solutions

Solution:
We represent the units with the digit [tex]u[/tex].
The hundreds digit is [tex]3u[/tex], therefore, the tens digit is the average: [tex]\frac{u + 3u}{2} = 2u[/tex].
The number has the form of the digits [tex](3u, 2u, u)[/tex].
Since they are distinct digits, [tex]3u \le 9[/tex].
Therefore, [tex]u[/tex] can be [tex]1[/tex], [tex]2[/tex], or [tex]3[/tex].
(If [tex]u = 0[/tex] the number does not have [tex]3[/tex] distinct digits).
If [tex]u = 1[/tex]: Digits [tex](3, 2, 1) \rightarrow 321[/tex].
Sum of digits [tex]= 6[/tex] (not divisible by [tex]9[/tex]).
If [tex]u = 2[/tex]: Digits [tex](6, 4, 2) \rightarrow 642[/tex].
Sum of digits [tex]= 12[/tex] (not divisible by [tex]9[/tex]).
If [tex]u = 3[/tex]: Digits [tex](9, 6, 3) \rightarrow 963[/tex].
Sum of digits [tex]= 18[/tex] (divisible by [tex]9[/tex]).
There is only one number that meets all the conditions: [tex]963[/tex].

Solution:
The initial ratio is [tex]\frac{12}{18} = \frac{2}{3}[/tex].
Let [tex]x[/tex] be the number of marbles of each color added.
The new ratio is: [tex]\frac{12 + x}{18 + x} = \frac{4}{5}[/tex].
Solving the equation: [tex]5(12 + x) = 4(18 + x)[/tex]
[tex]\Rightarrow 60 + 5x = 72 + 4x[/tex]
[tex]\Rightarrow x = 12[/tex].
[tex]12[/tex] marbles of each color were added.
Verification: New ratio is [tex]\frac{12 + 12}{18 + 12} = \frac{24}{30} = \frac{4}{5}[/tex].

Solution:
All angles of a rectangle measure [tex]90^\circ[/tex].
All the angles of an equilateral triangle measure [tex]60^\circ[/tex].
A joining vertex has a [tex]90^\circ[/tex] angle (from the rectangle) and a [tex]60^\circ[/tex] angle (from the triangle) adjacent to it along the [tex]6[/tex] cm side.
The total angle is the sum of the two adjacent angles: [tex]90^\circ + 60^\circ = 150^\circ[/tex].

Solution:
We consider positive integers less than [tex]50[/tex], which is the set [tex]\{1, 2, ..., 49\}[/tex].
Total is [tex]49[/tex] numbers.
Numbers divisible by [tex]2[/tex] (even numbers): [tex]49 \div 2 = 24.5 \rightarrow 24[/tex] numbers.
Numbers divisible by [tex]3[/tex]: [tex]49 \div 3 = 16.33 \rightarrow 16[/tex] numbers.
Numbers divisible by both (multiples of [tex]6[/tex]): [tex]49 \div 6 = 8.16 \rightarrow 8[/tex] numbers.
Using the Inclusion-Exclusion Principle, numbers divisible by [tex]2[/tex] or [tex]3 = 24 + 16 - 8 = 32[/tex].
Numbers NOT divisible by [tex]2[/tex] or [tex]3 = 49 - 32 = 17[/tex].

Solution:
The net is the flat development of all the faces of the 3D figure.
The total area of the net is equal to the total surface area of the cube.
A cube has [tex]6[/tex] identical square faces.
The area of one face is: [tex]5 \text{ cm} \times 5 \text{ cm} = 25 \text{ cm}^2[/tex].
The total area [tex]= 6 \times 25 \text{ cm}^2 = 150 \text{ cm}^2[/tex].