answer:1. Let ( x = z_1 ) and ( y = z_2 ). We are given that ( |z_1| = |z_2| = 1 ) and ( {z_1, z_2} neq {-i, i} ). The equation to solve is: [ z_1(z_1^3 + z_1^2 + rz_1 + 1) = z_2(z_2^3 + z_2^2 + rz_2 + 1). ] This can be rewritten as: [ x^4 + x^3 + rx^2 + x = y^4 + y^3 + ry^2 + y. ] 2. Adding 1 to both sides, we get: [ x^4 + x^3 + rx^2 + x + 1 = y^4 + y^3 + ry^2 + y + 1. ] This can be factored as: [ x^2 left( x^2 + frac{1}{x^2} + x + frac{1}{x} + r right) = y^2 left( y^2 + frac{1}{y^2} + y + frac{1}{y} + r right). ] 3. Since ( |x| = |y| = 1 ), ( frac{1}{x} ) and ( frac{1}{y} ) are the conjugates of ( x ) and ( y ) respectively. Thus, the quantities in parentheses are real. We can conclude that either the quantities in parentheses are zero or ( frac{y^2}{x^2} in mathbb{R} ). 4. Given ( |y| = |x| = 1 ), this implies ( y = pm ix ) or ( y = pm x ). The case ( y = x ) is prohibited by the problem statement. Hence, we try ( y = -x ). 5. Substituting ( y = -x ) into the original equation gives: [ x^3 + x = -x^3 - x, ] which simplifies to: [ 2x(x^2 + 1) = 0. ] The roots of this equation are also prohibited by the problem statement. Thus, we consider ( y = pm ix ). 6. Since ( y = -ix Longleftrightarrow x = iy ), we may assume without loss of generality that ( y = ix ). Substituting this into the original equation gives: [ x^3 + rx^2 + x = -ix^3 - rx^2 + ix, ] which simplifies to: [ (1 + i)x^2 + 2rx + (1 - i) = 0. ] 7. Solving for ( r ) gives: [ r = -frac{(1 + i)x^2 + (1 - i)}{2x} = -left( frac{(1 + i)x}{2} + frac{1 - i}{2x} right). ] This can be rewritten as: [ r = -frac{1}{2} left( x + frac{1}{x} + i left( x - frac{1}{x} right) right). ] 8. The real part of ( x ) on the unit circle is ( frac{1}{2} left( x + frac{1}{x} right) ) and the imaginary part is ( frac{1}{2i} left( x - frac{1}{x} right) ). Thus, our expression becomes: [ r = Im(x) - Re(x). ] This implies: [ -sqrt{2} leq r leq sqrt{2}. ] 9. Alternatively, we can have: [ x^2 + frac{1}{x^2} + x + frac{1}{x} + r = 0, ] for some two distinct ( x ). This can be rewritten as: [ r = -2(Re(x^2) + Re(x)). ] Letting ( x = e^{itheta} ), we seek to minimize the function ( -2(cos{2theta} + cos{theta}) ), which gives us the bounds ( -4 leq r leq frac{9}{4} ). 10. Since ( -4 ) occurs only at ( theta = 0 ) and only ( x = 1 ) has that argument, there are not two distinct ( x ) which are solutions for ( r = -4 ). Thus, ( -4 < r leq frac{9}{4} ). 11. Since ( [-1, 1] ) falls entirely within that interval, our answer is: [ |a| + |b| = 4 + frac{9}{4} = frac{25}{4}. ] Therefore, the final answer is: [ 100p + q = 100 times 25 + 4 = 2504. ] The final answer is (boxed{2504}).

answer:Using the trigonometric identity, [(1 + tan x)(1 + tan (30^circ - x)) = 1 + tan x + 1 - tan x = 2,] where x and 30^circ - x are the angles that sum to 30^circ. Let us pair the terms from x = 1^circ to x = 15^circ: begin{align*} (1 + tan 1^circ)(1 + tan 29^circ) &= 2, (1 + tan 2^circ)(1 + tan 28^circ) &= 2, &dots, (1 + tan 15^circ)(1 + tan 15^circ) &= 2. end{align*} This gives us 15 pairs, each equal to 2. Thus, the product simplifies to [(1 + tan 1^circ)(1 + tan 2^circ)(1 + tan 3^circ) dotsm (1 + tan 30^circ) = 2^{15}.] So, n = boxed{15}.

answer:1. **Identify the series and its properties:** The given series is (2, 5, 8, 11, 14, ldots, 50). This is an arithmetic series where the first term (a = 2) and the common difference (d = 3). 2. **Find the number of terms in the series:** To find the number of terms (n), we use the formula for the (n)-th term of an arithmetic series: [ a_n = a + (n-1)d ] Given (a_n = 50), we solve for (n): [ 50 = 2 + (n-1) cdot 3 ] [ 50 = 2 + 3n - 3 ] [ 50 = 3n - 1 ] [ 51 = 3n ] [ n = frac{51}{3} = 17 ] So, there are 17 terms in the series. 3. **Calculate the sum of the series:** The sum (S_n) of the first (n) terms of an arithmetic series is given by: [ S_n = frac{n}{2} left(2a + (n-1)dright) ] Substituting (n = 17), (a = 2), and (d = 3): [ S_{17} = frac{17}{2} left(2 cdot 2 + (17-1) cdot 3right) ] [ S_{17} = frac{17}{2} left(4 + 16 cdot 3right) ] [ S_{17} = frac{17}{2} left(4 + 48right) ] [ S_{17} = frac{17}{2} cdot 52 ] [ S_{17} = 17 cdot 26 ] [ S_{17} = 442 ] 4. **Verify the result:** The sum of the series is indeed 442. (boxed{442})

answer:To determine the probability that the note "MAME" remains on the top after Vasya folds and then randomly positions the paper, we need to consider the structure of the paper and the possible configurations it may end up in. 1. **Understanding the Paper Folding**: - Initially, the paper is folded in quarters, creating a note with "MAME" written on one of the quadrants. - When unfolded, the paper has 4 quadrants on one side and 4 quadrants on the other, totaling 8 quadrants. 2. **Possible Positions**: - The paper has 2 sides, each with 4 quadrants. - Since the folding and positioning are random, any of the 8 quadrants can be on top with equal probability. 3. **Calculating the Probability**: - The total number of possible quadrants that can appear on top is 8. - The "MAME" note occupies exactly 1 of these 8 quadrants. 4. **Formulating the Probability**: [ text{Probability} = frac{text{Number of favorable outcomes}}{text{Total number of outcomes}} = frac{1}{8} ] # Conclusion: The probability that the note "MAME" remains on top after unfolding and randomly positioning the paper is (boxed{frac{1}{8}}).