Math Olympiad Problems for Grades 7 and 8: Problems with Solutions

Solution:
Let [tex]N[/tex] be the number of coins.
The problem states that [tex]N[/tex] leaves a remainder of [tex]3[/tex] when divided by [tex]4[/tex], and a remainder of [tex]4[/tex] when divided by [tex]5[/tex].
Notice that in both cases, the arrangement is exactly [tex]1[/tex] coin short of completing a full stack.
This means that if [tex]1[/tex] more coin were added ([tex]N + 1[/tex]), the number would be perfectly divisible by both [tex]4[/tex] and [tex]5[/tex].
The smallest common multiple of [tex]4[/tex] and [tex]5[/tex] is [tex]20[/tex].
Therefore, [tex]N + 1 = 20 \Rightarrow N = 19[/tex].

Solution:
We solve this using the fundamental counting principle by choosing colors for the panels from left to right.
For the 1st panel, any of the [tex]3[/tex] colors can be chosen ([tex]3[/tex] options).
For the 2nd panel, any color except the one used for the 1st panel can be chosen ([tex]2[/tex] options).
For the 3rd panel, any color except the one used for the 2nd panel can be chosen ([tex]2[/tex] options).
For the 4th panel, any color except the one used for the 3rd panel can be chosen ([tex]2[/tex] options).
Total number of ways [tex]= 3 \times 2 \times 2 \times 2 = 24[/tex].

Solution:
First, calculate the total circumference ([tex]C[/tex]) of the circular track: [tex]C = 2\pi r = 2 \times 3 \times 30 = 180[/tex] meters.
The rover travels one-third of this circumference: [tex]\frac{180}{3} = 60[/tex] meters.
Average speed is defined as total distance divided by total time: Speed [tex]= \frac{60 \text{ m}}{40 \text{ s}} = 1.5 \text{ m/s}[/tex].

Solution:
The wall, the ground, and the ladder form a right-angled triangle.
The cosine of the angle [tex]\theta[/tex] is the ratio of the adjacent side (distance from the wall) to the hypotenuse (the length of the ladder, [tex]L[/tex]).
We have [tex]\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{5}{L}[/tex].
We are given that [tex]\cos(\theta) = 0.5 = \frac{1}{2}[/tex].
Therefore, [tex]\frac{5}{L} = \frac{1}{2}[/tex], which implies [tex]L = 5 \times 2 = 10[/tex] meters.

Solution:
Let the original two-digit number be [tex]10a + b[/tex], where [tex]a[/tex] is the tens digit and [tex]b[/tex] is the units digit.
The reversed number is [tex]10b + a[/tex].
According to the problem, the reversed number minus the original number is [tex]36[/tex]: [tex](10b + a) - (10a + b) = 36[/tex]
[tex]\Rightarrow 9b - 9a = 36[/tex]
[tex]\Rightarrow b - a = 4[/tex].
We also know the sum of the digits is [tex]10[/tex]: [tex]a + b = 10[/tex].
Adding the two equations: [tex]2b = 14 \Rightarrow b = 7[/tex].
Substituting back: [tex]a + 7 = 10 \Rightarrow a = 3[/tex].
The original password is [tex]a[/tex] followed by [tex]b[/tex], which is [tex]37[/tex].