MaxWong

The problem setter has made a mistake. Cosine of an angle must be within -1 and 1. -15/3 is -5, which is outside of the range. (Unless the angle is a complex number, of course, but that is out of consideration for, supposedly, a math problem in high school.)

You can use Chinese Remainder Theorem.

The problem translates to: $\begin{cases}1000 \leq N \leq 9999\\N \equiv 5 \pmod 9\\N \equiv 3 \pmod 7\\ N \equiv 3 \pmod 5\end{cases}$

Now, we let $N = 9k + 5$ for some integer k. Since $N \equiv 3 \pmod 7$,

$\begin{array}{rclc}9k+5&\equiv&3&\pmod7\\ 2k&\equiv&-2&\pmod{7}\\ k&\equiv&6&\pmod7\end{array}$

Then we let $k = 7n + 6$ for some integer n. Then $N = 9(7n + 6) + 5 = 63n + 59$.

Repeat this process one more time with N = 3 (mod 5), and you will get $N = 315m + 248$ for some integer m.

We can try m = 0, 1, 2, 3, ... to see which one gives the smallest possible 4-digit number. You will get $N = 315(3) + 248 = 1193$. You are more than welcome to check that this solution satisfies the conditions described in the problem.

The quadratic polynomial has two distinct real roots. That means the discriminant must be positive.

$\Delta > 0\\ m^2 - 4(1)(m) > 0\\ m^2 - 4m > 0\\ m > 4\text{ or }m < 0$

Can you write that inequality in interval notation?

Suppose the first two digits are 6's.

The only possibilities are 660, 661, ..., 666, 667, ..., 669.

You can count the other possibilities by hand. Just make sure that you don't count 666 three times and 0 cannot be the first digit, and you're good to go.

b divides 168 and c divides 490 by definition of gcd.

$\gcd(b, c) \geq \gcd(168, 490) = 14$

Therefore the smallest value is 14.

Using chain rule, 

$\begin{array}{rcl}f(x) &=& (2x - 1)^5\\ f'(x) &=& 5(2x - 1)^4 \dfrac{d}{dx} (2x - 1)\\ &=& 5(2x - 1)^4 (2)\\ &=& 10(2x - 1)^4\\ f'(2) &=& 10(2(2) - 1)^4\\ &=& 810\end{array}$

If f(x) has a real number value, 

$\dfrac{x - 7}{4 - x} \geq 0\\ (x - 7)(4 - x) \geq 0\text{ and }x \neq 4\\ 4 < x \leq 7$

The smallest possible integer in that range is x = 5. 

If you have a polynomial, substitute 1 into the polynomial and that will be the sum of the coefficients.

Therefore, sum of coefficients = -(3 - 1)(1 + 2(3 - 1) - 7(1 - 5)) = -66.

To check the answer, we can go the normal way and see if we really get this sum of coefficients.

-(3 - c)(c + 2(3 - c) - 7(c - 5))

= (c - 3)(c + (6 - 2c) - (7c - 35))

= (c - 3)(c + 6 - 2c - 7c + 35)

= (c - 3)(-8c + 41)

= -8c^2 + 24c + 41c - 123

= -8c^2 + 65c - 123

sum of coefficients = -8 + 65 - 123 = -66.

Therefore the answer is correct.

Let $y = a\cdot b^x +c$ for some constants a, b, and c

Substitute (x, y) = (0, 900) (x, y) = (5, 200) and (x, y) = (4, 450) to get the system of simultaneous equations $\begin{cases}a + c= 900\\a\cdot b^4 + c=450\\a\cdot b^5 + c=200\end{cases}$.

It is really hard to find the exact solutions of this system of equations, but the approximate solution is (a, b, c) = (-147.7, 1.418, 1047.75).

Here's the graph:

heureka posted a great solution, so I will post a different one.

$\quad1 - 2 + 3 - 4 + \cdots + 2019 - 2020 + 2021\\ =(1 - 2) + (3 - 4) + \cdots + (2019 - 2020) + 2021\\ = \underbrace{(-1) + (-1) + \cdots + (-1)}_{1010\text{ times}} + 2021\\ =-1010 + 2021\\ = 1011$