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因子問題-組合與排列
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Given that 2310 = 2 × 3 × 5 × 7 × 11. (A) Including 1 and 2310, how many different factors does 2310 have? Ans: 2? = 32 Solution: Since 2310 = 21 × 31 × 51 × 71 × 111, any factor of 2310 must be of the form 2^a × 3^b × 5^c × 7^d × 11^e, where a ∈ {0, 1}, b ∈ {0, 1}, c ∈ {0, 1}, d ∈ {0, 1}, e ∈ {0, 1}, that is, each of a, b, c, d, e can take value either 0 or 1. Note that the smallest factor of 1 can be obtained when a = b = c = d = e = 0. Also, the largest factor of 2310 can be obtained when a = b = c = d = e = 1. Since there are two choices (0 or 1) of value for each of a, b, c, d, e, the number of all the factors of 2310 is 2 × 2 × 2 × 2 × 2 = 2? = 32. (B) Using Prime Factor Multiplication with Index Notation, express the product of all factors of 2310 (including 1 and 2310). Ans: 21? × 31? × 51? × 71? × 111? Solution: Following (a), any factor of 2310 must be of the form 2^a × 3^b × 5^c × 7^d × 11^e, where a ∈ {0, 1}, b ∈ {0, 1}, c ∈ {0, 1}, d ∈ {0, 1}, e ∈ {0, 1}. The product of all factors of 2310 is the product of all 32 different numbers in the form 2^a × 3^b × 5^c × 7^d × 11^e where the combination of {a, b, c, d, e} takes different values. For every fixed value of a (0 or 1), the number of cases for the variation in {b, c, d, e} is 2? = 16. Therefore, out of the all 32 numbers, 16 of them would have the part 2? = 1 while the other 16 of them would have the part 21 = 2. Similarly, there are 16 factors which have 3? = 1 while the other 16 have 31 = 3. There are 16 factors which have 5? = 1 while the other 16 have 51 = 5. There are 16 factors which have 7? = 1 while the other 16 have 71 = 7. There are 16 factors which have 11? = 1 while the other 16 have 111 = 11. As a result, the product of all the 32 factors is 21? × 31? × 51? × 71? × 111?.
其他解答:63D0B758E2D502CC
因子問題-組合與排列
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Give 2310=2*3*5*7*11 A. Including 1 and 2310, how many different factors does 2310 have ? Ans is 2^5=32 B. Using Prime Factor Multiplication with Index Notation, express the product of all factors of 2310(including 1 and 2310) Ans is 2^16*3^16*5^16*7^16*11^16 不明白怎樣 做到這答案最佳解答:
Given that 2310 = 2 × 3 × 5 × 7 × 11. (A) Including 1 and 2310, how many different factors does 2310 have? Ans: 2? = 32 Solution: Since 2310 = 21 × 31 × 51 × 71 × 111, any factor of 2310 must be of the form 2^a × 3^b × 5^c × 7^d × 11^e, where a ∈ {0, 1}, b ∈ {0, 1}, c ∈ {0, 1}, d ∈ {0, 1}, e ∈ {0, 1}, that is, each of a, b, c, d, e can take value either 0 or 1. Note that the smallest factor of 1 can be obtained when a = b = c = d = e = 0. Also, the largest factor of 2310 can be obtained when a = b = c = d = e = 1. Since there are two choices (0 or 1) of value for each of a, b, c, d, e, the number of all the factors of 2310 is 2 × 2 × 2 × 2 × 2 = 2? = 32. (B) Using Prime Factor Multiplication with Index Notation, express the product of all factors of 2310 (including 1 and 2310). Ans: 21? × 31? × 51? × 71? × 111? Solution: Following (a), any factor of 2310 must be of the form 2^a × 3^b × 5^c × 7^d × 11^e, where a ∈ {0, 1}, b ∈ {0, 1}, c ∈ {0, 1}, d ∈ {0, 1}, e ∈ {0, 1}. The product of all factors of 2310 is the product of all 32 different numbers in the form 2^a × 3^b × 5^c × 7^d × 11^e where the combination of {a, b, c, d, e} takes different values. For every fixed value of a (0 or 1), the number of cases for the variation in {b, c, d, e} is 2? = 16. Therefore, out of the all 32 numbers, 16 of them would have the part 2? = 1 while the other 16 of them would have the part 21 = 2. Similarly, there are 16 factors which have 3? = 1 while the other 16 have 31 = 3. There are 16 factors which have 5? = 1 while the other 16 have 51 = 5. There are 16 factors which have 7? = 1 while the other 16 have 71 = 7. There are 16 factors which have 11? = 1 while the other 16 have 111 = 11. As a result, the product of all the 32 factors is 21? × 31? × 51? × 71? × 111?.
其他解答:63D0B758E2D502CC
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