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$$\triangle ABC$$ is right-angled triangle, $$\angle ACB = 90^\circ$$. The sides are of integral length. If the product of the legs is equal to the sum of all sides, then find the sides of the triangle. How many such triangle are there?
Let the legs be $$a = BC$$ and $$b = AC$$, so the hypotenuse is $$c = \sqrt{a^2 + b^2}$$ and the condition is $$ab = a + b + \sqrt{a^2 + b^2}$$. Squaring $$(ab - a - b)^2 = a^2 + b^2$$ gives $$a^2b^2 - 2a^2b - 2ab^2 + 2ab = 0$$, that is $$ab - 2a - 2b + 2 = 0$$. Adding 2 to both sides gives $$(a - 2)(b - 2) = 2$$. Negative factor pairs make a side zero, so $$\{a - 2, b - 2\} = \{1, 2\}$$, the legs are 3 and 4, and the hypotenuse is 5 (check $$3 \times 4 = 12 = 3 + 4 + 5$$). Hence only one triangle, with sides 3, 4, 5, is possible (2 if the ordered assignments $$(BC, AC) = (3, 4)$$ and $$(4, 3)$$ are counted separately).
$$n$$ is a natural number and $$n \le 2026$$. If $$n$$ divides $$20^n + 17k$$, for some natural number $$k$$, then find the number of such $$n$$.
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In triangle $$ABC$$, the incircle, with center $$I$$, touches the side $$BC$$ at point $$D$$. Line $$DI$$ meets $$AC$$ at $$P$$. The tangent line from $$P$$ to the incircle (different from $$AC$$) intersects $$AB$$ at $$Q$$. If line $$QI$$ and $$BC$$ intersect at point $$R$$, prove that quadrilateral $$AIRC$$ is a cyclic.
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Given a sequence $$a_1, a_2, a_3, \ldots$$, where $$a_n = a_{n-1} - a_{n-2}$$, for $$n \in \mathbb{N}$$ and $$n \ge 3$$. Also, $$S_n = a_1 + a_2 + a_3 + \cdots + a_{n-1} + a_n$$. If $$S_{2003} = 2026$$ and $$S_{2026} = 2003$$, then find $$S_{2030}$$.
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Find all possible values of natural numbers $$m$$ and $$n$$ such that $$12mn - 175 = 105m + 140n$$.
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Find all natural numbers $$x, y, z$$ such that $$\sqrt{\frac{2026}{x+y}} + \sqrt{\frac{2026}{y+z}} + \sqrt{\frac{2026}{z+x}}$$ is also a natural number.
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How many ways can you mark 8 squares of an $$8 \times 8$$ chessboard so that no two marked squares are in the same row or column, and none of the four corner squares is marked? (Rotations and reflections are considered different.)
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In a triangle $$ABC$$, we have $$\angle ACB = \angle BAC + 90^\circ$$. The point $$D$$ is taken on the extended $$BC$$ such that $$AC = AD$$. A point $$E$$ is taken such that points $$A$$ and $$E$$ are on the opposite sides of side $$BC$$ and $$\angle EBC = \angle BAC$$, $$\angle EDC = \frac{1}{2}\angle BAC$$. Prove that $$\angle CED = \angle ABC$$.
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