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

Solution:
The divisibility rule for [tex]9[/tex] states that a number is divisible by [tex]9[/tex] if and only if the sum of its digits is divisible by [tex]9[/tex].
The sum of the known digits is [tex]3 + 5 + 6 = 14[/tex].
We want to find a digit ”[tex]A[/tex]” (from [tex]0[/tex] to [tex]9[/tex]) such that [tex]14 + A[/tex] is a multiple of [tex]9[/tex].
The closest multiple of [tex]9[/tex] greater than [tex]14[/tex] is [tex]18[/tex].
If [tex]14 + A = 18[/tex], then [tex]A = 4[/tex].
The next multiple is [tex]27[/tex], requiring [tex]A = 13[/tex], which is not a single digit.

Solution:
The arithmetic mean is the sum of the numbers divided by the count.
The total sum of the four numbers is [tex]15 \times 4 = 60[/tex].
The sum of the three known numbers is [tex]10 + 12 + 18 = 40[/tex].
The fourth number is [tex]60 - 40 = 20[/tex].

Solution:
The width of the rectangle is [tex]4[/tex] cm and the length is three times its width ([tex]4 \times 3 = 12[/tex] cm).
The area of the rectangle is [tex]12 \times 4 = 48 \text{ cm}^2[/tex].
Since squares have a side length equal to the width of the rectangle ([tex]4[/tex] cm), the area of one square is [tex]4 \times 4 = 16 \text{ cm}^2[/tex].
The total area is the area of the rectangle plus the area of the two squares: [tex]48 + 16 + 16 = 80 \text{ cm}^2[/tex].

Solution:
The sum of the interior angles of a triangle is [tex]180^\circ[/tex].
Let [tex]x[/tex] be the measure of the third angle.
Then, the second angle measures [tex]2x[/tex].
Equation: [tex]45^\circ + 2x + x = 180^\circ[/tex]
[tex]\Rightarrow 45^\circ + 3x = 180^\circ[/tex]
[tex]\Rightarrow 3x = 135^\circ[/tex]
[tex]\Rightarrow x = 45^\circ[/tex].
The three angles are [tex]45^\circ[/tex], [tex]90^\circ[/tex], and [tex]45^\circ[/tex].
The largest angle measures [tex]90^\circ[/tex].

Solution:
We identify the rule of the logical pattern between the terms: [tex](2 \times 2) + 1 = 5[/tex], [tex](5 \times 2) + 1 = 11[/tex], [tex](11 \times 2) + 1 = 23[/tex], [tex](23 \times 2) + 1 = 47[/tex].
The rule is ”multiply the previous term by [tex]2[/tex] and add [tex]1[/tex]”.
Applying the rule to the last term: [tex](47 \times 2) + 1 = 95[/tex].

Solution:
This is a permutations problem without repetition.
For the first letter, we have [tex]4[/tex] options.
For the second letter, we have [tex]3[/tex] options left.
For the third letter, we have [tex]2[/tex] options.
Multiplying the options: [tex]4 \times 3 \times 2 = 24[/tex] different codes.

Solution:
Let [tex]x[/tex] be the total capacity of the tank.
Initially, it contains [tex]\frac{3}{5}x[/tex].
After adding [tex]12[/tex] liters, it contains [tex]\frac{3}{4}x[/tex].
Equation: [tex]\frac{3}{5}x + 12 = \frac{3}{4}x[/tex].
Simplifying: [tex]12 = \frac{3}{4}x - \frac{3}{5}x[/tex]
[tex]\Rightarrow 12 = \frac{15x - 12x}{20}[/tex]
[tex]\Rightarrow 12 = \frac{3x}{20}[/tex]
[tex]\Rightarrow 240 = 3x[/tex]
[tex]\Rightarrow x = 80[/tex] liters.

Solution:
Numbers divisible by [tex]12[/tex] are multiples of [tex]12[/tex].
List of multiples starting with [tex]12[/tex]: [tex]\{12, 24, 36, 48, 60, 72, 84, 96\}[/tex].
The next multiple is [tex]108[/tex] (greater than [tex]100[/tex]).
Counting the numbers in the list, there are [tex]8[/tex] whole numbers between [tex]10[/tex] and [tex]100[/tex] that are divisible by [tex]12[/tex].

Solution:
Using the inclusion-exclusion principle, calculate the total number of students who play at least one sport: (play Soccer) + (play Basketball) - (play Both) [tex]= 18 + 15 - 7 = 26[/tex].
There are [tex]26[/tex] students playing sports.
The number of students who do not play either is: [tex]30 - 26 = 4[/tex].

Solution:
The perimeter of both figures is [tex]40[/tex] cm.
For the square ([tex]4[/tex] sides): Side [tex]= 40 \div 4 = 10[/tex] cm.
For the equilateral triangle ([tex]3[/tex] sides): Side [tex]= \frac{40}{3}[/tex] cm.
The absolute difference is: [tex]\frac{40}{3} - 10 = \frac{40}{3} - \frac{30}{3} = \frac{10}{3}[/tex] cm.