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question:Billy's mom sends him to get ketchup. She gives him 10 and tells him to get the best deal on ketchup that he can and to spend all 10 on ketchup. He finds a bottle with 10 oz that cost 1 each. He finds a bottle that costs 2 that contains 16 ounces. He finds a bottle with 25 ounces that costs 2.5. He finds a 5 bottle that contains 50 ounces. Finally, he finds a 10 bottle with 200 ounces. How many bottles of ketchup does he buy?
answer:To determine the best deal on ketchup per ounce and how many bottles Billy can buy with 10, we calculate the cost per ounce for each option: 1. For the 10 oz bottle costing 1: - Cost per ounce = frac{1}{10 text{ oz}} = 0.10 per ounce. 2. For the 16 oz bottle costing 2: - Cost per ounce = frac{2}{16 text{ oz}} = 0.125 per ounce. 3. For the 25 oz bottle costing 2.5: - Cost per ounce = frac{2.5}{25 text{ oz}} = 0.10 per ounce. (Note: There seems to be a mistake in the original solution; it should be frac{2.5}{25 text{ oz}} = 0.10 per ounce.) 4. For the 50 oz bottle costing 5: - Cost per ounce = frac{5}{50 text{ oz}} = 0.10 per ounce. 5. For the 200 oz bottle costing 10: - Cost per ounce = frac{10}{200 text{ oz}} = 0.05 per ounce. The 200 ounce bottle is the cheapest per ounce at 0.05 compared to the other options. Since Billy has 10 to spend and the 200 oz bottle costs 10, he can buy exactly 1 bottle of this size. Therefore, Billy buys boxed{1} bottle of ketchup.
question:Find the matrix mathbf{N} that swaps the rows of a matrix. In other words, [mathbf{N} begin{pmatrix} a & b c & d end{pmatrix} = begin{pmatrix} c & d a & b end{pmatrix}.] If no such matrix mathbf{N} exists, then enter the zero matrix.
answer:Let mathbf{N} = begin{pmatrix} p & q r & s end{pmatrix}. Then [mathbf{N} begin{pmatrix} a & b c & d end{pmatrix} = begin{pmatrix} p & q r & s end{pmatrix} begin{pmatrix} a & b c & d end{pmatrix} = begin{pmatrix} pa + qc & pb + qd ra + sc & rb + sd end{pmatrix}.] We want this to be equal to begin{pmatrix} c & d a & b end{pmatrix}. Setting up equations: - pa + qc = c and pb + qd = d - ra + sc = a and rb + sd = b Assuming a = 1, b = 0, c = 0, d = 1 (for simplicity and without loss of generality): - From pa + qc = 0 and pb + qd = 1, we get p cdot 1 + q cdot 0 = 0 and p cdot 0 + q cdot 1 = 1, hence p = 0 and q = 1. - From ra + sc = 1 and rb + sd = 0, we get r cdot 1 + s cdot 0 = 1 and r cdot 0 + s cdot 1 = 0, hence r = 1 and s = 0. Thus, mathbf{N} = begin{pmatrix} 0 & 1 1 & 0 end{pmatrix}. Checking: [mathbf{N} begin{pmatrix} a & b c & d end{pmatrix} = begin{pmatrix} 0 & 1 1 & 0 end{pmatrix} begin{pmatrix} a & b c & d end{pmatrix} = begin{pmatrix} c & d a & b end{pmatrix}.] This matches exactly with the desired outcome, so the matrix mathbf{N} that swaps the rows is boxed{begin{pmatrix} 0 & 1 1 & 0 end{pmatrix}}.
question:f the line y = 2 intersects the parabola y = x^2 + kx 1 at only one point, then ( ) A: k = 2 B: k = 2 C: k = pm 2 D: The value of k cannot be determined
answer:Substitute y = -2 into y = x^2 + kx - 1, we get x^2 + kx + 1 = 0 According to the condition that they intersect at only one point, we have Delta = k^2 - 4 = 0, Thus, k = pm 2. Therefore, the correct option is boxed{C}.
question:Let ( n geq 2 ) be a positive integer. A subset of positive integers ( S ) is said to be comprehensive if for every integer ( 0 leq x < n ), there is a subset of ( S ) whose sum has remainder ( x ) when divided by ( n ). Note that the empty set has sum 0. Show that if a set ( S ) is comprehensive, then there is some (not necessarily proper) subset of ( S ) with at most ( n1 ) elements which is also comprehensive.
answer:We need to show that if ( S ) is a comprehensive set, then there exists a subset of ( S ) with at most ( n-1 ) elements that is also comprehensive. 1. **Initial Setup:** Consider ( S = {s_1, s_2, ldots, s_k} ) where ( k geq n ). Since ( S ) is comprehensive, by definition, for every integer ( 0 leq x < n ), there is a subset of ( S ) whose sum has remainder ( x ) when divided by ( n ). 2. **Process and Tracking Remainders:** We start with the empty set, which has a sum of 0 and thus, the set of remainders, ( T ), initially is ( {0} ). As we add each element ( s_i ) from ( S ), we update the set of remainders. 3. **Updating Remainders:** For each element ( s_i ) added, the set of possible remainders becomes: [ T cup {(t + s_i) mod n mid t in T} ] This means the new remainders include all previous remainders, plus the sum of each previous remainder plus ( s_i ), taken modulo ( n ). 4. **Final Remainders Set:** At the beginning of the process, we have: [ T_1 = {0} ] After adding elements one by one, since ( S ) is comprehensive, the final set of remainders must be: [ {0, 1, ldots, n-1} ] 5. **Element Contribution and Bin Size Constraint:** Since ( |S| = k geq n ), we add in at least ( n ) elements. During this process, because ( |{0, 1, ldots, n-1}| = n ), by the Pigeonhole Principle, there must be at least one element ( s_i ) in ( S ) whose addition does not introduce any new remainders. 6. **Removing Non-Contributing Element:** Let's denote the comprehensive set of remainders before adding ( s_i ) by ( T ). If adding ( s_i ) does not change the set of remainders, then: [ T = T cup {(t + s_i) mod n mid t in T } ] implies ( T ) remained unchanged. This indicates that ( S ) remains comprehensive without ( s_i ). 7. **Repeating the Process:** By iteratively removing such non-contributing elements ( s_i ), we reduce the set size while maintaining its comprehensive property. We repeat this process until we've reduced the set to have at most ( n-1 ) elements. **Conclusion:** Thus, we have shown that from any comprehensive set ( S ) of size at least ( n ), it is possible to find a comprehensive subset of ( S ) containing at most ( n-1 ) elements. [ boxed{} ]