answer:If Megan drank 3 out of her 17 bottles of water, she would have: 17 - 3 = boxed{14} bottles left.

answer:From the equation xy + z = yz + x = zx + y = 56, we proceed similarly. 1. Rewrite the first equality: [ xy + z - yz - x = 0 implies x(y-1) - y(z-1) = 0. ] Similarly, from the second: [ yz + x - zx - y = 0 implies y(z-1) - z(x-1) = 0. ] 2. Assume x = y. Substitute into xy + z = 56: [ x^2 + z = 56. ] Also, from yz + x = 56 with y = x: [ xz + x = 56 implies x(z+1) = 56. ] 3. As 56 = 7 times 8, let's try x = 7. Then z + 1 = 8 implies z = 7. Substitute in x^2 + z = 56: [ 7^2 + 7 = 49 + 7 = 56. ] Therefore, x = y = z = 7. So, x+y+z = 7+7+7 = boxed{21}.

answer:Define the modified Fibonacci sequence as: [ G_1 = 2, quad G_2 = 3, quad G_n = G_{n-1} + G_{n-2} text{ for } n geq 3. ] Compute this sequence modulo 10: [ begin{align*} G_3 & = G_2 + G_1 = 3 + 2 = 5, G_4 & = G_3 + G_2 = 5 + 3 = 8, G_5 & = G_4 + G_3 = 8 + 5 = 13 equiv 3 pmod{10}, G_6 & = G_5 + G_4 = 3 + 8 = 11 equiv 1 pmod{10}, G_7 & = G_6 + G_5 = 1 + 3 = 4, G_8 & = G_7 + G_6 = 4 + 1 = 5, G_9 & = G_8 + G_7 = 5 + 4 = 9, G_{10} & = G_9 + G_8 = 9 + 5 = 14 equiv 4 pmod{10}, G_{11} & = G_{10} + G_9 = 4 + 9 = 13 equiv 3 pmod{10}, G_{12} & = G_{11} + G_{10} = 3 + 4 = 7, G_{13} & = G_{12} + G_{11} = 7 + 3 = 10 equiv 0 pmod{10}, G_{14} & = G_{13} + G_{12} = 0 + 7 = 7, G_{15} & = G_{14} + G_{13} = 7 + 0 = 7, G_{16} & = G_{15} + G_{14} = 7 + 7 = 14 equiv 4 pmod{10}, G_{17} & = G_{16} + G_{15} = 4 + 7 = 11 equiv 1 pmod{10}, G_{18} & = G_{17} + G_{16} = 1 + 4 = 5, G_{19} & = G_{18} + G_{17} = 5 + 1 = 6. end{align*} ] All digits from 0 to 9 have now appeared in the units position of this modified sequence. The last digit to appear is 6. Conclusion: The last digit to appear in the units position of a number in this modified sequence is 6. The final answer is boxed{textbf{(C)} 6}.

answer:# Detailed Solution: 1. **Initial Configuration:** - Suppose we place the circles in a basic grid layout within an ( n text{cm} ) square such that each circle touches its neighbor and the sides of the square. - Given that the diameter of each circle is ( 1 text{cm} ), we can place exactly ( n ) circles in one row. - There are ( n ) such rows in the ( n ) cm square. - Therefore, in this arrangement, the total number of circles is ( n times n = n^2 ). 2. **Alternate Arrangement:** - To increase the number of circles, we can try a different arrangement. - In the second row, we place the circles in the gaps between the circles of the first row. This requires calculating the vertical distance between the centers of two vertically touching circles. - The distance calculation: - Consider the centers of three circles forming an equilateral triangle with sides of (1 text{cm} ). - The height ( h ) of this triangle (altitude) is given by ( frac{sqrt{3}}{2} ). - The vertical distance between two adjacent rows is: [ 2 - left( 1 + frac{sqrt{3}}{2} right) = frac{4 - sqrt{3}}{2} = frac{4 - 2sqrt{3}}{2} + 1 = 1 + frac{2 - sqrt{3}}{2} = 1 - frac{sqrt{3}}{2} ] 3. **Determining ( n ) for Increased Density:** - We need to calculate the number of rows that can fit within the square of side ( n ) under this new arrangement: - Each pair of rows will fit together more tightly than the initial arrangement because the height reduction allows for additional rows. - For any number ( n ) of rows, there will be ( n-1 ) reduced height intervals. - We need to satisfy the inequality: [ (n-1) left(frac{2 - sqrt{3}}{2}right) geq frac{sqrt{3}}{2} ] 4. **Solving the Inequality:** - Simplify by multiplying both sides by 2: [ (n-1) (2 - sqrt{3}) geq sqrt{3} ] - Divide both sides by (2 - sqrt{3}): [ n - 1 geq frac{sqrt{3}}{2 - sqrt{3}} ] - Approximating ( frac{sqrt{3}}{2 - sqrt{3}} approx 2.732 ): [ n - 1 geq 2.732 implies n geq 3.732 implies n geq 8 ] 5. **Verifying and Counting Circles:** - For ( n = 8 ): - The effective arrangement will fit 5 full rows of 8 circles and 4 full rows of 7 circles. - Therefore, the total number of circles is: [ 5 times 8 + 4 times 7 = 40 + 28 = 68 ] - Since ( 68 > 8^2 = 64 ), our configuration yields more than ( n^2 ) circles. # Conclusion: Thus, we find that it is indeed possible to configure the circles such that their number exceeds ( n^2 ) for ( n ) sufficiently large. Specifically: [ boxed{text{Yes}} ]