#### Problem 1 [***]

There are unique integers \(a_2, a_3, ... a_7 \) such that

$$\frac{5}{7} = \frac{a_2}{2!} + \frac{a_3}{3!} + \frac{a_4}{4!} + \frac{a_5}{5!} + \frac{a_6}{6!} + \frac{a_7}{7!}$$

where \( 0 <= a_i < i\) for i = 2,3...,7

Find \(a_2, a_3 ... a_7 \)

[AHSME]

1. Remove the fractions.

2. Look at this as a remainder problem first what's the remainder when divided by 7 etc.

2. Look at this as a remainder problem first what's the remainder when divided by 7 etc.

#### Problem 2 [*]

Prove: log(1 + 2 + 3) = log (1) + log(2) + log(3)

[@mathematicsprof]

Remember the log rules.

#### Problem 3 [**]

$$(6x + 5)^2(3x +2)(x+1) = 35$$

[@mathematician_3]

1. Substitution is the key \((6x + 5)^2 = 36x^2 + 60x + 25 = 12(3x^2 + 5x + 2) + 1 = 12(3x + 2)(x + 1) + 1\)

#### Problem 4 [*]

The perimeter of the rectangle ABCD is 28 units E is the midpoint of the side AC and the two arc are tangent to each other. What are the dimensions of the rectangle?

[@five_triangles originally]

Find the implicit triangle and then apply the Pythagorean theorem.

#### Problem 5 [**]

This is a pair for comparison:[@matematik_man]

[@eylem_99]

Both depend on forming 30-60-90 triangles. Look especially at the square root 3 side.

#### Problem 6[**]

[UWSIM program]

1. Take logs to get rid of the exponents i.e. \( \log{x} \cdot \log{3} = \log{y} \cdot \log{4} \)

2. Then you can use substitution for log x or log y.

3. Its hard to see the simplification so it might be useful to substitute things in but essentially this reduces to \( \log{x} \cdot (\frac{\log{4}^2 - \log{3}^2}{\log{4}}) = \log{3}^2 - \log{4}^2 \)

4. The difference of squares terms cancel out and you're left with \( \log{x} = -\log{4} \)

2. Then you can use substitution for log x or log y.

3. Its hard to see the simplification so it might be useful to substitute things in but essentially this reduces to \( \log{x} \cdot (\frac{\log{4}^2 - \log{3}^2}{\log{4}}) = \log{3}^2 - \log{4}^2 \)

4. The difference of squares terms cancel out and you're left with \( \log{x} = -\log{4} \)

#### Problem 7 [**]

A,F,B,E,C on circle O. AC=BD=4, CG=3, AC//FE.

Find the area of AEF

[@five_triangles]

1. Note ACG is a 3-4-5 and BEG is similar to it.

2. Since EF is parallel to AC triangle DEG is also similar to the 3-4-5

3. Set x = DG then EG = 5/3X but x + 4 = 5/3(EG) . Solve and DG = x = 9/4.

4. Drop and altitude from A to EF. This divides into a 3-4-5 similar triangle where you have one side AE and a triangle similar to ABC where you also know all the sides.

5. This gives the altitude and base for AEF.

#### Problem 8 [***]

Let p be a 50 digit prime number, when squared its remainder when divided by 120 is

**not**1. What is \(p^2\mod 120\) ?[AOPS]

1. Factor 120

2. Observe the behavior of squares mod each of the factors.

3. If you're prime then you can never be 0 mod any of the factors.

4. Only 2 cases left.

2. Observe the behavior of squares mod each of the factors.

3. If you're prime then you can never be 0 mod any of the factors.

4. Only 2 cases left.

#### Problem 9 [*]

1. Find a polynomial with integer coefficients that has \(3 + \sqrt{5} \) as one root.2. Find a polynomial with integer coefficients that has \(\sqrt{5} + \sqrt{7} \) as one root.

3. Find a polynomial with integer coefficients that has \(3 + \sqrt{5} + \sqrt{7} \) as one root.

[Mike Lawler]

Use conjugates or square the sides enough to remove the radicals. Also don't forget about shifting a function to the right or left.

#### Problem 10 [*]

Find the smallest positive integer N such that \( \frac{N - 13}{5N + 6} \) is a reducible fraction.[AHSME]

Try using the Euclidean Algorithm to find the gcd between the numerator and denominator.

#### Problem 11 [*]

Find the value of \( \sqrt{100 \cdot 101 \cdot 102 \cdot 103 + 1} \) then find a general formula.

[@matematik_man]

There are a variety of methods that work. Start by looking at much smaller examples i.e. n = 1, 2 etc. Curve fit, use a linear system assume the expression is a square etc.

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