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Lemma: If \(x + \dfrac1x = 2\cos \alpha\), then \(x^n + \dfrac1{x^n} = 2\cos n\alpha\)

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\(a\equiv b^{-1}\pmod n\\ a\cdot b\equiv b^{-1}\cdot b \pmod n\\ ab \equiv 1 \pmod n\)

The required remainder is 1.

\(x^3 + y^3 = (x + y)(x^2 - xy + y^2) = (\color{blue}x + y\color{black})((\color{blue}x + y\color{black})^2\, - \,3\color{red}xy\color{black})\)

Now you can substitute the values.

Let me prove the "only if" part first.

\(x + 9y \equiv 0 \pmod {19}\\ 9(x + 9y) \equiv 0 \pmod {19}\\ 9x + 81y \equiv 0 \pmod{19}\\ 9x + 19(4y) + 5y \equiv 0 \pmod{19}\\ 9x + 5y \equiv 0 \pmod{19}\)

The "if" part would just be the above, but writing the steps in reverse order.

Thanks!

Notice that 

\(\dfrac1p = \dfrac{q}{pq}\\ \dfrac1q = \dfrac{p}{pq}\)

If we rewrote the fractions like this, then we could add the two fractions.

\(\dfrac{1}p + \dfrac1q = \dfrac{q}{pq} + \dfrac{p}{pq} = \dfrac{p + q}{pq} \)

Substituting pq = 9/2,

\(p + q = \dfrac92\)

Subtracting q on both sides,

\(p = \dfrac92 - q\)

Substituting this into pq = 9/2,

\(q\left(\dfrac92 - q\right) = \dfrac92\\ q^2 - \dfrac92 q + \dfrac92 = 0\\ 2q^2 - 9q + 9 = 0\)

You can solve this equation by using the formula \(q = \dfrac{-b\pm\sqrt{b^2- 4ac}}{2a}\), where a is the coefficient of x^2, b is the coefficient of x, and c is the constant term.

Thank you Dragan :)

We consider the discriminant of the left hand side.

Because the quadratic function is always positive, the discriminant is negative.

\((2(4K - 1))^2 - 4(15K^2 - 2K - 7) < 0\\ 64K^2 - 32K + 4 - 60K^2 + 8K + 28 < 0\\ 4K^2 - 24K + 32 < 0\\ K^2 - 6K + 8 < 0\\ \)

Solving gives \(2 < K < 4\). I will leave the solving process to you.

Thanks!

Divisible by 42 means all of below: 

1) divisible by 2

2) divisible by 3

3) divisible by 7

So, by 1), we know that x is an even digit.

By 2), the digit sum must be divisible by 3 because of the divisibility trick.

The digit sum is \(x + 3 + 5 + 7 + 9 + x = 2x + 24\)

We know that 24 is divisible by 3. That means if 2x is divisible by 3, then the entire number is divisible by 3.

We don't even need 3) to work out the answer.

Out of all even digits(0,2,4,6,8), there is only one digit satisfying the conditions, which is 6.

Adding gives

\((a + b + c)(1 + x + y) = 7(1 + x + y)\)

which means a + b + c = 7 if \(x + y \neq -1\).

(The formula is easily verifiable by expanding the terms and comparing to what you have after adding the three equations.)

An easy-to-understand solution:

After writing the terms, let S be the original sum and consider S/3.

Subtracting gives:

Now 2S/3 is the sum of geometric series 1/3 + 1/9 + 1/27 + ...

\(\dfrac{2S}3 = \dfrac{\dfrac13}{1 - \dfrac13} = \dfrac12\\ S = \dfrac34\)

Therefore the required sum is 3/4.

listfor(n, 0, 10, (11+2*n)) = (11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31) = 11 terms

A fun solution:

\(\displaystyle\sum_{k = 1}^n 1 = n\)

Keeping that in mind:

\(\displaystyle \sum_{n = 1}^\infty \frac{n}{3^n} = \sum_{n=1}^\infty \sum_{k=1}^n 3^{-n}\)

By Fubini's theorem, the sums are interchangeable.

\(\displaystyle \sum_{n = 1}^\infty \frac{n}{3^n} = \sum_{k=1}^\infty \sum_{n=k}^\infty 3^{-n} = \sum_{k =1 }^\infty \dfrac{3^{-k}}{1 - \dfrac13} = \dfrac32 \cdot \dfrac{\dfrac13}{1 - \dfrac13} = \dfrac34\)

Thanks!!

Let's look at the differences in speed each hour.....

hour

1      2    3   4   5    6    7   8    9    10

diff

-4    -3   -2  -1  0     1   2   3    4      5

so at hour 5 they are the same speed....then the originally slower train starts gaining so that at hour 9 it has made up the miles it lost in the first 4 hours of going slower

at hour 9 they should be at the same point

check:   9 hr (40 km/hr) = 360 km        

                          other train  36  + 37 +38 +39 +40 +41 +42 + 43 + 44  = 360 km

I understand now. Thank you.

sumfor(n, 1, 667, (3*n - 1) =667,667 + 2002 - 2003 =667,666