text: Quandaries and Queries Name: Simon Who is asking: Student Level of the question: Secondary Question: determine smallest positive integer that is divisible by each of the first ten counting numbers Hi Simon, I'm going to find the smallest positive integer N that is divisible by each of the first six counting numbers. The first six counting numbers are 1, 2, 3, 4, 5 and 6 but since every number is divisible by 1 I am only going to work with 2, 3, 4, 5 and 6. The first step is to find the prime decompositions. 2 = 21 3 = 31 4 = 2 2 = 22 5 = 51 6 = 21 31 The primes that appear in these decompositions are 2, 3 and 5. Since each of these counting numbers divide N, the three primes 2, 3 and 5 must divide N. Since these are the only primes that divide these counting numbers and since I want N to be as small as possible, the only primes that divide N are 2, 3 and 5. Thus the prime factorization of N is N = 2? 3? 5? All that is left is to determine what valued go in for the question marks. I want these values to be as small as possible to keep N small. 5 only appears once in the factorizations above (5 = 51) the power of 5 in the factorization of N is 1. N = 2? 3? 51 3 appears twice in the factorizations, (3 = 31 and 6 = 21 31) but in each case the power of 3 is 1 so the power of 3 in the prime factorization of N is 1. N = 2? 31 51 Finally, 2 appears 3 times and in one of these (4 = 22 ) the power of 2 is 2. Thus if this number is to divide N the power of 2 in the prime factorization of N must be at least 2 also. Since we want N as small as possible a power of 2 will suffice. Hence N = 22 31 51 = 60 Now you try the first ten counting numbers. Penny
explanation: The document is a response to a math question posed by a student named Simon. The question asks for the smallest positive integer that is divisible by each of the first ten counting numbers. The response provides a detailed explanation of how to find the smallest positive integer divisible by the first six counting numbers, using prime factorization. The document does not provide the answer to the original question but gives a method to solve a similar problem.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: Quandaries and Queries Name: Simon Who is asking: Student Level of the question: Secondary Question: determine smallest positive integer that is divisible by each of the first ten counting numbers Hi Simon, I'm going to find the smallest positive integer N that is divisible by each of the first six counting numbers. The first six counting numbers are 1, 2, 3, 4, 5 and 6 but since every number is divisible by 1 I am only going to work with 2, 3, 4, 5 and 6. The first step is to find the prime decompositions. 2 = 21 3 = 31 4 = 2 2 = 22 5 = 51 6 = 21 31 The primes that appear in these decompositions are 2, 3 and 5. Since each of these counting numbers divide N, the three primes 2, 3 and 5 must divide N. Since these are the only primes that divide these counting numbers and since I want N to be as small as possible, the only primes that divide N are 2, 3 and 5. Thus the prime factorization of N is N = 2? 3? 5? All that is left is to determine what valued go in for the question marks. I want these values to be as small as possible to keep N small. 5 only appears once in the factorizations above (5 = 51) the power of 5 in the factorization of N is 1. N = 2? 3? 51 3 appears twice in the factorizations, (3 = 31 and 6 = 21 31) but in each case the power of 3 is 1 so the power of 3 in the prime factorization of N is 1. N = 2? 31 51 Finally, 2 appears 3 times and in one of these (4 = 22 ) the power of 2 is 2. Thus if this number is to divide N the power of 2 in the prime factorization of N must be at least 2 also. Since we want N as small as possible a power of 2 will suffice. Hence N = 22 31 51 = 60 Now you try the first ten counting numbers. Penny
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a response to a math question posed by a student named Simon. The question asks for the smallest positive integer that is divisible by each of the first ten counting numbers. The response provides a detailed explanation of how to find the smallest positive integer divisible by the first six counting numbers, using prime factorization. The document does not provide the answer to the original question but gives a method to solve a similar problem.
  
  Label: question-and-answer