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let the resulting integer equal x. reword it another way, \(29x\equiv x(\text{mod }10^n)\rightarrow28x\equiv 0(\text{mod }10^n)\)where n is just some positive integer.

the prime factorization of 28 is 2*2*7, meaning that there must be at least a 5 in the prime factorization of x to multiply with one of the 2's in 28(for its last digit to equal zero and for n to be 1). If we were to plug 5 in, we see that it doesn't work, meaning that x must be at least 2 digits. That means that x must have at least 2 5's in its prime factorization for it to multiply with the 2 2's in 28(for its last 2 digits to equal zero and for n to be 2). Testing out 5*5=25, we can see that it works, and we are sure it's the smallest possible value, so the answer is \(25*29=\boxed{725}\)

the complex number -4 has a degree of 180 degrees. that means that the degrees of the solutions are going to be 180/4=45, 180/4+360/4=135, 180/4+360/4+360/4=225, and 180/4+360/4+360/4+360/4=315.

The magnitude of -4 is obviously 4, so the magnitude of the solutions are going to be 4^(1/4) = sqrt(2)

Therefore, the solutions in polar form are: \(\sqrt{2}(\cos(45)+i\sin(45)), \sqrt{2}(\cos(135)+i\sin(135)), \sqrt{2}(\cos(225)+i\sin(225)), \sqrt{2}(\cos(315)+i\sin(315))\)

Simplifying gives \(\boxed{-1-i, -1+i, 1-i, 1+i}\)

(notice that I gave a general way to solve this problem, but obviously, for this problem, you could factor it and solve it in a quicker way.)

The divisibility rule for 9 says that: \(S(n)\equiv n (\text{mod 9})\) 

This rule is true every number can be written as \(x_1+10x_2+...10^{n-1}x_n\), where \(x_n\) is the nth digit of the number, and since 10^n is always congruent to 1 mod 9 (because 10^n-1 is 99...9, which is divisible by 9), the expression reduces to \(x_1+1x_2+...1x_n\), which is simply the sum of the digits.

Anyway, back to the question, the original expression reduces to

 \(n \equiv n+n (\text{mod 9})\\n\equiv 0 (\text{mod 9})\)

Now, we just need to find the number of 3 digit multiples of 9. Can you finish it?

The distance from a point on the graph to the origin can be expressed as, in terms of x, as \(\sqrt{x^2+(x^2-3)^2}=\sqrt{x^4-6x^2+9+x^2}=\sqrt{x^4-5x^2+9}\)

the quartic inside the radical is just a quadratic in terms of x^2, so we can rewrite it in vertex form:

\(\sqrt{x^4-5x^2+9}=\sqrt{x^4-5x^2+(\frac{5}{2})^2+9-(\frac{5}{2})^2}=\sqrt{(x^2-\frac{5}{2})^2+\frac{11}{4}}\)

since \((x^2-2.5)^2\)will always be nonnegative for real numbers, the minimum value for \(\sqrt{(x^2-2.5)^2+\frac{11}{4}}\) is just equal to \(\boxed{\frac{\sqrt{11}}{2}}\) (specifically when x is equal to \(\pm\sqrt{2.5}\))

P.S. my method seems like total overkill for this problem so if anyone has an easier solution, please post it

let y = \(\sqrt[3]{x}\):

\(4y-\frac{2y^3}{y^2}=7+y\\4y-2y=7+y\\y=7\)

\(7=\sqrt[3]{x}\\x=343\)

i'm guessing that you multiplied 3*5*7*9*11 and then subtracted 1, but notice that 3 is a factor of 9, so the answer would be 5*7*9*11-1 = 3464

If there are 22 students who bought their calculators to class, and 13 of them are girls, then that means that 9 of them are boys. Since there are 16 boys in total, there are 16-9=7 boys who did not bring their calculators to class.

Idk if this is right or not, but this is my best shot at it:

Since 1 is a constant, I would treat this as a binomial:

\(((a+b+c+d)+1)^N\\=x_1\cdot(a+b+c+d)^N+x_2\cdot(a+b+c+d)^{N-1}+...x_{N}\cdot(a+b+c+d)+1\)

, where \(x_1, x_2, .... x_{N}\) are just random coefficients that are irrelevant, since we only care about the number of terms.

By the multinomial theorem (the theorem doesn't exactly state this but it will suffice for our problem), the number of terms in \((a+b+c+d)^{N}\) is equal to the number of ways to split a chunk of length N into 4 smaller nonnegative integer lengths (since there are 4 terms).

Since the problem specifies that all terms must include all variables a, b, c, and d, we can assume that the chunk N already includes at least 1 of a, b, c, and d, so we can say that the number of terms in \((a+b+c+d)^{N}\) with all variables included is equal to the number of ways to split a chunk of length N-4 into 4 smaller nonnegative integer lengths. (Notice that for N=1, 2, and 3, the value is negative, which makes sense, since\(a+b+c+d,(a+b+c+d)^2,(a+b+c+d)^3\)

all don't include any terms that include all variables.)

By stars and bars (the nonnegative version), there are exactly \({5+(N-4)-1 \choose 5-1}={N \choose 4}\), since 5 bars is needed to split the N-4 chunk to split it into 4 smaller chunks.

Notice that 14 choose 4 is 1001, so it is the answer (I won't go too in-depth because I have already written so much)

Edit: Made some adjustments to my answer, but I just noticed that there are so much better and faster answers online.

Applying Euclidean's algorithm, we get:

\(\gcd(3n+4, n) \\=\gcd(3n+4-3n, n)\\=\gcd(4, n)\)

The factors of 4 are 1, 2, and 4. Sum them up and you'll get the answer.

First, simplify the first equation:

\(x=1+\frac{\sqrt{3}}{x}\\x^2-x=\sqrt{3}\)

Then, notice that \(\frac{1}{(x+1)(x-2)}\) simplifies to \(\frac{1}{x^2-x-2}\), so we can replace it like so:

\(\frac{1}{\sqrt{3}-2}\)

To simplify, just multiply the top and bottom by the conjugate

\(\frac{\sqrt{3}+2}{(\sqrt{3}+2)(\sqrt{3}-2)}\\=\frac{\sqrt{3}+2}{3-4}=\frac{2+\sqrt{3}}{-1}\)

\(|2|+|3|+|-1|=\boxed{6}\)