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question:A 5digit positive integer is defined as cute if its n digits (where n=5) are an arrangement of the set {1, 2, 3, 4, 5} and its first k digits form an integer that is divisible by k, for k = 1, 2, ..., 5. Determine how many cute 5digit integers exist. A) 0 B) 1 C) 2 D) 3 E) 4
answer:1. **Constraints Analysis**: - The integer abcde must satisfy: - Digits are a permutation of {1, 2, 3, 4, 5}. - Each k-digit leading segment (for k = 1, 2, ..., 5) must be divisible by k. 2. **Divisibility by 1**: - Any number is divisible by 1, so a can be any digit from 1 to 5. 3. **Divisibility by 5**: - The number abcde must end in a digit divisible by 5. Here, e must be 5, as that is the only choice. 4. **Divisibility by 4**: - abcd (last two digits cd) must be divisible by 4. Valid pairs (from remaining digits {1, 2, 3, 4}) are {12, 32, 24}. 5. **Divisibility by 2**: - ab must be divisible by 2, so b must be even. Possible choices for b are 2 and 4. 6. **Divisibility by 3**: - The sum of digits in abc must make it divisible by 3. Selection of digits for a, b, c should consider this condition. 7. **Checking Valid Combinations** using cases from step 4: - If cd = 12, then c = 1 and d = 2. We need abc divisible by 3, but no suitable ab since e=5. - If cd = 24, then c = 2 and d = 4, leaving ab3e. The suitable one would be ab = 13 so a = 3, b = 1. - If cd = 32, then c = 3 and d = 2, leaving ab4e. The suitable one would be ab = 14 so a = 1, b = 4. 8. **Conclusion**: - The viable cute 5-digit integers are 31245 and 41325. 2 The final answer is boxed{C) 2}
question:A YouTube video is 100 hours long. Lila decides to watch it at a certain speed. Roger, her friend, also watches the same video at the average speed. They both watched six of the same videos, and the total number of hours they watched is 900. What is the ratio of Lila's watching speed to the average speed?
answer:Let's denote Lila's watching speed as L and the average speed as A. Since they both watched six videos, we can set up the following equation based on the total number of hours they watched: 6 * 100/L + 6 * 100/A = 900 We can simplify this equation by dividing everything by 6: 100/L + 100/A = 150 Now, let's multiply everything by L * A to get rid of the denominators: 100A + 100L = 150LA Now, we know that Roger watches at the average speed, so A = 1 (since watching at average speed means watching at the normal 1x speed). Let's substitute A with 1: 100(1) + 100L = 150L 100 + 100L = 150L Now, let's solve for L: 150L - 100L = 100 50L = 100 L = 100/50 L = 2 So, Lila watches at twice the average speed. Therefore, the ratio of Lila's watching speed to the average speed is boxed{2:1} .
question:w many continuous paths from A to B, along segments of the figure, do not revisit any of the seven labeled points? [asy] draw((0,0)(3,0)(3,2)(0,2)(0,0)cycle,linewidth(2)); draw((0,2)(1,0)(3,2)(0,2)cycle,linewidth(2)); draw((0,2)(1.5,3.5)(3,2),linewidth(2)); draw((1.5,3.5)(1,0),linewidth(2)); label(A,(1.5,3.5),N); label(B,(0,0),SW); label(C,(0,2),W); label(D,(3,2),E); label(E,(3,0),SE); label(F,(1,0),S); label(G,(1.5,1),S); [/asy]
answer:We denote a path from A to B by writing the labeled points visited, such as A-C-B (first going to C then to B). Case 1: Path ends in C-B. - Direct path: A-C-B - Via D: A-D-C-B - Via D and F: A-D-F-C-B - Via D, E, and F: A-D-E-F-C-B - Via G: A-G-C-B - Via G and F: A-G-F-C-B Case 2: Path ends in F-B. - Direct paths: A-F-B, A-G-F-B - Via C: A-C-F-B, A-C-D-F-B, A-C-D-E-F-B - Via D: A-D-F-B, A-D-E-F-B Counting these, we have 6 paths in Case 1 and 7 paths in Case 2, giving a total of boxed{13} such paths.
question:A real number ( x ) is randomly chosen between 100 and 200. If ([sqrt{x}] = 12), what is the probability that ([sqrt{100 x}] = 120)? (A) (frac{2}{25}). (B) (frac{241}{2500}). (C) (frac{1}{10}). (D) (frac{96}{625}). (E) 1.
answer:To solve this problem, we need to determine the probability that a randomly chosen real number ( x ) between 100 and 200 satisfies the given conditions. Here are the steps involved: 1. **Determine the Range for ( x ) from ( [sqrt{x}]=12 ):** By definition, [ [sqrt{x}]=12 implies 12 leq sqrt{x} < 13 ] Squaring the inequality, [ 12^2 leq x < 13^2 implies 144 leq x < 169 ] 2. **Determine the Range for ( x ) from ( [sqrt{100x}]=120 ):** By definition, [ [sqrt{100x}]=120 implies 120 leq sqrt{100x} < 121 ] Dividing the inequality by 10, [ 12 leq 10sqrt{x} < 12.1 ] Simplifying, [ 12 leq 10sqrt{x} < 12.1 implies frac{12}{10} leq sqrt{x} < frac{12.1}{10} implies 1.2 leq sqrt{x} < 1.21 ] Squaring each part of the inequality, [ (1.2)^2 leq x < (1.21)^2 implies 1.44 leq x < 1.4641 ] 3. **Combining Both Conditions:** The probabilities for ( x ) are constrained by both conditions: [ 12 leq sqrt{x} < 13 quad text{and} quad 12 leq sqrt{x} < 12.1 ] The valid range for ( x ) must satisfy both constraints, so we take the intersection of these ranges: [ 12 leq sqrt{x} < 12.1 implies 144 leq x < 146.41 ] 4. **Calculate the Probability:** The length of the interval for ( x ) satisfying both conditions is: [ 12.1^2 - 12^2 = 146.41 - 144 = 2.41 ] The overall length of the interval for ( x ) when ( 12 leq sqrt{x} < 13 ) is: [ 13^2 - 12^2 = 169 - 144 = 25 ] Therefore, the probability is given by: [ p = frac{2.41}{25} = frac{241}{2500} ] Conclusion: The probability that a randomly selected ( x ) between 100 and 200 satisfies ( [sqrt{x}]=12 ) and ( [sqrt{100x}]=120 ) is: [ boxed{frac{241}{2500}} ] Thus, the correct answer is ((B)).