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question:Brody's calculator can run for some hours on a full battery. Brody has used three quarters of its battery up, and he is about to take a twohour math exam that will require his calculator the whole time. After the exam, Brody's calculator will have 13 hours of battery left. How many hours can Brody's calculator run on a full battery?

answer:If Brody has used three quarters of the battery, he has one quarter of the battery left. After the two-hour exam, he will have 13 hours of battery left, which means that one quarter of the battery lasts for 13 + 2 = 15 hours. Since one quarter of the battery lasts for 15 hours, the full battery would last for 15 hours * 4 = 60 hours. Therefore, Brody's calculator can run for boxed{60} hours on a full battery.

question:f x, y, z > 0, find the smallest possible value of [leftlfloor{frac{x+y}{z}}rightrfloor+leftlfloor{frac{y+z}{x}}rightrfloor+leftlfloor{frac{z+x}{y}}rightrfloor.]

answer:Since lfloor{x}rfloor geq x - 1 for all x, we have [ Biglfloor{frac{x+y}{z}}Bigrfloor + Biglfloor{frac{y+z}{x}}Bigrfloor + Biglfloor{frac{z+x}{y}}Bigrfloor geq frac{x+y}{z} + frac{y+z}{x} + frac{z+x}{y} - 3. ] Applying the AM-GM inequality to each fraction pair: [ frac{x}{y} + frac{y}{x} geq 2, quad frac{y}{z} + frac{z}{y} geq 2, quad frac{z}{x} + frac{x}{z} geq 2. ] Thus, the expression simplifies to: [ frac{x+y}{z} + frac{y+z}{x} + frac{z+x}{y} geq 6. ] Subtracting 3 from each side: [ Biglfloor{frac{x+y}{z}}Bigrfloor + Biglfloor{frac{y+z}{x}}Bigrfloor + Biglfloor{frac{z+x}{y}}Bigrfloor geq 3. ] Since it is an integer, the smallest possible value is boxed{4}. This can be achieved by selecting specific values for x, y, and z such as (x, y, z) = (6, 8, 9).

question:the diagram, AOB is a sector of a circle with center O and angle AOB=90^circ. OY is drawn perpendicular to AB and intersects AB at X. Determine the length of XY if the radius of the circle is 10 units. [asy] draw((0,0)(10,0),black+linewidth(1)); draw((0,0)(7.07,7.07)..(10,0)..(7.07,7.07)(0,0),black+linewidth(1)); draw((7.07,7.07)(7.07,7.07),black+linewidth(1)); label(O,(0,0),W); label(A,(7.07,7.07),N); label(B,(7.07,7.07),S); label(X,(7.07,0),SE); label(Y,(10,0),E); label(10,(0,0)(7.07,7.07),NW); label(10,(0,0)(7.07,7.07),SW); [/asy]

answer:Since OY is a radius of the circle, we have OY = 10. To find XY, we first need to determine the length of OX. Given that angle AOB = 90^circ, the triangle triangle OAB is isosceles with OA = OB = 10. The angle angle OAB = frac{1}{2}(180^circ - 90^circ) = 45^circ. Therefore, angle OAX = 180^circ - 45^circ - 90^circ = 45^circ, making triangle OAX a 45^circ-45^circ-90^circ triangle. In a 45^circ-45^circ-90^circ triangle, the legs are equal, and each leg is frac{1}{sqrt{2}} times the hypotenuse, which is OA. Thus, OX = frac{10}{sqrt{2}} = 5sqrt{2}. Hence, XY = OY - OX = 10 - 5sqrt{2}. Conclusion: The length of XY is boxed{10 - 5sqrt{2}} approx 2.93.

question:a 100 times 25 rectangular grid, each cell contains a nonnegative real number. The number in the ith row and jth column is denoted as x_{i, j} (i = 1, 2, ldots, 100; j = 1, 2, ldots, 25). Then, the numbers in each column are rearranged in descending order from top to bottom to form a new grid with entries x^{prime}_{i, j} such that x^{prime}_{1, j} geq x^{prime}_{2, j} geq cdots geq x^{prime}_{100, j} for j = 1, 2, ldots, 25. Find the smallest natural number k such that if the entries in the original grid satisfy sum_{j=1}^{25} x_{i, j} leq 1 for all i = 1, 2, ldots, 100, then for i geq k, it holds that sum_{j=1}^{25} x^{prime}_{i, j} leq 1 in the rearranged grid.

answer:To solve the problem, we need to find the smallest natural number ( k ) such that for any ( i geq k ), the condition (sum_{j=1}^{25} x_{i, j}^{prime} leqslant 1) holds after reordering the elements as specified. 1. **Analysis of the given conditions**: - Each element ( x_{i, j} ) is a non-negative real number. - We are given that (sum_{j=1}^{25} x_{i, j} leq 1) for all rows ( i = 1,2, ldots, 100 ) in the original table. 2. **Understand the reordering process**: - When the values ( x_{i, j} ) are reordered in descending order within each column in the second table, it holds that ( x^{prime}_{1, j} geq x^{prime}_{2, j} geq cdots geq x^{prime}_{100, j} ) for each ( j = 1,2, ldots, 25 ). 3. **Consider a specific choice of ( x_{i, j} )**: - Let us choose ( x_{i, j} ) as follows: x_{i,j} = begin{cases} 0, & text{if } 4(j-1) + 1 leq i leq 4j frac{1}{24}, & text{otherwise} end{cases} for each ( j = 1,2, ldots, 25 ). 4. **Check the summation condition in the original table**: - For this choice, each row sum becomes: sum_{j=1}^{25} x_{i, j} = 0 + 24 left(frac{1}{24}right) = 1 - Hence, the condition ( sum_{j=1}^{25} x_{i,j} leq 1 ) is satisfied for all ( i = 1, 2, ldots, 100 ). 5. **Analyze the reordering in the new table**: - After reordering the elements, we get: x_{i, j}^{prime} = begin{cases} frac{1}{24}, & text{if } 1 leq i leq 96 0, & text{if } 97 leq i leq 100 end{cases} for each ( j = 1,2, ldots, 25 ). 6. **Check for which ( k ) the condition holds**: - For ( i = 1 ) to ( i = 96 ): sum_{j=1}^{25} x_{i, j}^{prime} = 25 times frac{1}{24} = frac{25}{24} > 1 - For ( i = 97 ) to ( i = 100 ): sum_{j=1}^{25} x_{i, j}^{prime} = 25 times 0 = 0 leq 1 Therefore, the smallest value of ( k ) such that (sum_{j=1}^{25} x_{i, j}^{prime} leqslant 1) for all ( i geq k ) is 97. 7. **General argument for validation**: - Table 1 must have at least one row ( r ) such that every number ( x_{r, 1}, x_{r, 2}, ldots, x_{r, 25} ) appears in the top 97 rows of Table 2. - If the above condition does not hold, then it implies some number in the first row of Table 1 does not move to the top 97 rows in Table 2, leading to contradiction with ( 97 times 25 = 2425 ) positions needed. - Since ( x^{prime}_{i,j} leqslant x_{r,j} ) for ( i geq 97 ) and thus (sum_{j=1}^{25} x^{prime}_{i,j} leq sum_{j=1}^{25} x_{r,j} leq 1). Therefore, the smallest value of ( k ) is: [ boxed{97} ]