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If x is the product of the positive integers from 1 to 9, inclusive,
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25 May 2016, 04:27
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84% (00:54) correct 16% (01:29) wrong based on 100 sessions
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Re: If x is the product of the positive integers from 1 to 9, inclusive,
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25 May 2016, 04:54
x= 9! = 9* 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = (3^2) * (2^3) * 7 * (2*3) * 5 * 2^2 * 3 * 2 * 1 = 2^7 * 3^4 * 5 * 7 = 2^i * 3^k * 5^m * 7^p i+k+m+p = 7+4+1+1 = 13 Answer E
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If x is the product of the positive integers from 1 to 9, inclusive,
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31 May 2016, 15:27
Given x = 9! = \(2^i * 3^k * 5^m * 7^p\) Find the number of powers of prime numbers: \(n / i^1 + n / i^2 ...\) until i^z <n, z any positive number z=>1 Number of 2s in expression 9!: i = \(\frac{9}{2} + \frac{9}{8}≈ 5 + 1 = 6\) Number of 3s in expression 9!: k = \(\frac{9}{3} + \frac{9}{9} ≈ 3 + 1 = 4\) Number of 5s in expression 9!: m = \(\frac{9}{5} ≈ 2\) Number of 7s in expression 9!: p = \(\frac{9}{7} ≈ 1\) i + k + m + p = 6 + 4 + 2 + 1 = 13 Answer: E
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Re: If x is the product of the positive integers from 1 to 9, inclusive,
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31 May 2016, 20:42
fantaisie wrote: Given x = 9! = \(2^i * 3^k * 5^m * 7^p\)
Find the number of powers of prime numbers: \(n / i^1 + n / i^2 ...\) until i^z <n, z any positive number z=>1
Number of 2s in expression 9!: i = \(\frac{9}{2} + \frac{9}{8}≈ 5 + 1 = 6\)
Number of 3s in expression 9!: k = \(\frac{9}{3} + \frac{9}{9} ≈ 3 + 1 = 4\)
Number of 5s in expression 9!: m = \(\frac{9}{5} ≈ 2\)
Number of 7s in expression 9!: p = \(\frac{9}{7} ≈ 1\)
i + k + m + p = 6 + 4 + 2 + 1 = 13
Answer: E Hi fantaisie , It seems that you have missed 2^2 while trying to find powers of 2 in 9! . Also you have rounded up in a few cases . Hope this helps !! \(\frac{9}{2}+\frac{9}{4}+\frac{9}{8}= 4+2+1 = 7\) \(\frac{9}{3}+\frac{9}{9}= 3+1 = 4\) \(\frac{9}{5} = 1\) \(\frac{9}{7} = 1\) Finding the powers of a prime number p, in the n!The formula is: \(\frac{n}{p}+\frac{n}{p^2}+\frac{n}{p^3}+...+\frac{n}{p^k}\), where \(k\) must be chosen such that \(p^k\leq{n}\)
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Re: If x is the product of the positive integers from 1 to 9, inclusive,
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01 Jun 2016, 02:56
Skywalker18 wrote: fantaisie wrote: Given x = 9! = \(2^i * 3^k * 5^m * 7^p\)
Find the number of powers of prime numbers: \(n / i^1 + n / i^2 ...\) until i^z <n, z any positive number z=>1
Number of 2s in expression 9!: i = \(\frac{9}{2} + \frac{9}{8}≈ 5 + 1 = 6\)
Number of 3s in expression 9!: k = \(\frac{9}{3} + \frac{9}{9} ≈ 3 + 1 = 4\)
Number of 5s in expression 9!: m = \(\frac{9}{5} ≈ 2\)
Number of 7s in expression 9!: p = \(\frac{9}{7} ≈ 1\)
i + k + m + p = 6 + 4 + 2 + 1 = 13
Answer: E Hi fantaisie , It seems that you have missed 2^2 while trying to find powers of 2 in 9! . Also you have rounded up in a few cases . Hope this helps !! \(\frac{9}{2}+\frac{9}{4}+\frac{9}{8}= 4+2+1 = 7\) \(\frac{9}{3}+\frac{9}{9}= 3+1 = 4\) \(\frac{9}{5} = 1\) \(\frac{9}{7} = 1\) Finding the powers of a prime number p, in the n!The formula is: \(\frac{n}{p}+\frac{n}{p^2}+\frac{n}{p^3}+...+\frac{n}{p^k}\), where \(k\) must be chosen such that \(p^k\leq{n}\) Thank you very much for pointing out my mistake! I have one question: what's the rule when it comes to rounding in this formula? \(\frac{9}{5}\) = 1.8, do we only take into account actual integers? Thank you!
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Re: If x is the product of the positive integers from 1 to 9, inclusive,
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01 Jun 2016, 03:03
fantaisie wrote: Thank you very much for pointing out my mistake! I have one question: what's the rule when it comes to rounding in this formula? \(\frac{9}{5}\) = 1.8, do we only take into account actual integers? Thank you!
Hi, rounding of is always done to the lower integer... Reason  say we are looking for number of 5 in 9!  9/5 + 9/5^2 = 1.8 + 0.3 = 1+0 = 1.. so 1 in 1.8 talks of integer 5 in the product....
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Re: If x is the product of the positive integers from 1 to 9, inclusive,
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