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

Solution:
For a number to be divisible by [tex]36[/tex], it must be divisible by both [tex]4[/tex] and [tex]9[/tex].
Rule for [tex]4[/tex]: The last two digits [tex]b2[/tex] must be divisible by [tex]4[/tex], meaning [tex]b[/tex] can be [tex]1, 3, 5, 7[/tex], or [tex]9[/tex].
Rule for [tex]9[/tex]: The sum of all digits must be divisible by [tex]9[/tex].
The sum is [tex]5 + a + 7 + b + 2 = 14 + a + b[/tex].
To maximize [tex]a + b[/tex], we test the largest values for [tex]b[/tex]: If [tex]b = 9[/tex], sum is [tex]23 + a[/tex].
The next multiple of [tex]9[/tex] is [tex]27[/tex], so [tex]a = 4[/tex].
Sum [tex]a + b = 13[/tex].
If [tex]b = 7[/tex], sum is [tex]21 + a[/tex].
Multiple is [tex]27[/tex], so [tex]a = 6[/tex].
Sum [tex]a + b = 13[/tex].
If [tex]b = 5[/tex], sum is [tex]19 + a[/tex].
Multiple is [tex]27[/tex], so [tex]a = 8[/tex].
Sum [tex]a + b = 13[/tex].
The maximum sum is exactly [tex]13[/tex].

Solution:
The total number of paths from [tex](0,0)[/tex] to [tex](4,4)[/tex] without restrictions requires [tex]4[/tex] Right and [tex]4[/tex] Up moves ([tex]8[/tex] total).
Total paths [tex]= \binom{8}{4} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70[/tex].
A path goes through the anomaly if it goes from [tex](0,0)[/tex] to [tex](2,2)[/tex] AND from [tex](2,2)[/tex] to [tex](4,4)[/tex].
Paths to [tex](2,2) = \binom{4}{2} = 6[/tex].
Paths from [tex](2,2)[/tex] to [tex](4,4) = \binom{4}{2} = 6[/tex].
Total bad paths [tex]= 6 \times 6 = 36[/tex].
Safe paths = Total - Bad [tex]= 70 - 36 = 34[/tex].

Solution:
Evaluate the expression inside the parentheses first: [tex]3 \oplus 6 = \frac{3 \times 6}{3 + 6} = \frac{18}{9} = 2[/tex].
Substitute this back into the expression: [tex](3 \oplus 6) \oplus 2 = 2 \oplus 2[/tex].
Calculate the final operation: [tex]2 \oplus 2 = \frac{2 \times 2}{2 + 2} = \frac{4}{4} = 1[/tex].

Solution:
Using Heron’s Formula for a triangle with known sides [tex]a = 13[/tex], [tex]b = 14[/tex], [tex]c = 15[/tex]: Area [tex]= \sqrt{s(s - a)(s - b)(s - c)}[/tex], where semi-perimeter [tex]s = \frac{13 + 14 + 15}{2} = 21[/tex].
Substituting values: Area [tex]= \sqrt{21(21 - 13)(21 - 14)(21 - 15)} = \sqrt{21(8)(7)(6)}[/tex].
Factoring: [tex]\sqrt{(7 \times 3) \times (2^3) \times 7 \times (2 \times 3)} = \sqrt{7^2 \times 3^2 \times 2^4} = 7 \times 3 \times 4 = 84 \text{ m}^2[/tex].

Solution:
We find the repeating pattern of unit digits.
For powers of [tex]3[/tex]: [tex]3^1 \rightarrow 3[/tex], [tex]3^2 \rightarrow 9[/tex], [tex]3^3 \rightarrow 7[/tex], [tex]3^4 \rightarrow 1[/tex] (cycle of [tex]4[/tex]).
Dividing exponent [tex]2026[/tex] by [tex]4[/tex] yields a remainder of [tex]2[/tex].
The 2nd step in the pattern is [tex]9[/tex].
For powers of [tex]4[/tex]: [tex]4^1 \rightarrow 4[/tex], [tex]4^2 \rightarrow 6[/tex] (cycle of [tex]2[/tex]).
Odd powers end in [tex]4[/tex], even in [tex]6[/tex].
Since [tex]2025[/tex] is odd, [tex]4^{2025}[/tex] ends in [tex]4[/tex].
Multiplying the unit digits: [tex]9 \times 4 = 36[/tex].
The unit digit of the massive product is [tex]6[/tex].