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The "length" of a positive integer greater than 1 is the number of pri
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18 Nov 2014, 08:32
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Re: The "length" of a positive integer greater than 1 is the number of pri
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18 Nov 2014, 09:04
5950 =5x1190 =5x5x238 =5x5x2x118 =5x5x2x2x59
5 prime factors including repeats
Answer E!



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Re: The "length" of a positive integer greater than 1 is the number of pri
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18 Nov 2014, 22:08
5950=2*5*595 =2*5*5*119 =2*5*5*17*7
so , length is 5
Answer is E



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Re: The "length" of a positive integer greater than 1 is the number of pri
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19 Nov 2014, 00:09
Answer = E = 5 \(5950 = 2*5*595\) \(= 2 * 5 * 5 * 119\) \(= 2 * 5 * 5 * 7 * 17\)
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Re: The "length" of a positive integer greater than 1 is the number of pri
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19 Nov 2014, 00:11
DangerPenguin wrote: 5950 =5x1190 =5x5x238 =5x5x2x118 =5x5x2x2x59
5 prime factors including repeats
Answer E! Seems to be typo; though answer is correct
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Re: The "length" of a positive integer greater than 1 is the number of pri
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19 Nov 2014, 07:58
Official Solution:The "length" of a positive integer greater than 1 is the number of prime numbers, including repeats, that are factors of that integer. For instance, the length of 20 is 3, because \(2 \times 2 \times 5 = 20\). What is the length of 5,950?A. 1 B. 2 C. 3 D. 4 E. 5 In this problem, we should notice that the "length" of an integer is defined in some way involving prime factors. Therefore, even if we are not 100% sure what "length" means, we should factor 5,950 down to its prime factors and examine what we get. First, it's easy to take out a factor of 10: \(5950 = 10 \times 595\). 10 is very simple to factor: \(10 = 2 \times 5\). Since 595 ends in 5, we know that 595 is divisible by 5. Actually performing this division, we wind up with \(595 = 5 \times 119\). So \(5950 = 2 \times 5 \times 5 \times 119\). Now, 119 "looks" prime, but you must test it with primes up to the square root of 119, which is approximately 11. (If 119 is not prime, then at least one of its prime factors must be smaller than the square root of 119.) It turns out that 119 is divisible by 7. \(119 = 7 \times 17\). Thus, we now have the full prime factorization of 5,950: \(5950 = 2 \times 5 \times 5 \times 7 \times 17\). Finally, we return to the definition of "length." We are given the example that the length of 20 is 3, since \(2 \times 2 \times 5 = 20\). So we can see that "length" is just the number of prime numbers in the prime factorization, counting repeats (such as 2 in the example of 20). Thus, the length of 5,950 is 5. Answer: E.
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Re: The "length" of a positive integer greater than 1 is the number of pri
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19 Nov 2014, 07:59
Bunuel wrote: Official Solution:
The "length" of a positive integer greater than 1 is the number of prime numbers, including repeats, that are factors of that integer. For instance, the length of 20 is 3, because \(2 \times 2 \times 5 = 20\). What is the length of 5,950?
A. 1 B. 2 C. 3 D. 4 E. 5 In this problem, we should notice that the "length" of an integer is defined in some way involving prime factors. Therefore, even if we are not 100% sure what "length" means, we should factor 5,950 down to its prime factors and examine what we get. First, it's easy to take out a factor of 10: \(5950 = 10 \times 595\). 10 is very simple to factor: \(10 = 2 \times 5\). Since 595 ends in 5, we know that 595 is divisible by 5. Actually performing this division, we wind up with \(595 = 5 \times 119\). So \(5950 = 2 \times 5 \times 5 \times 119\). Now, 119 "looks" prime, but you must test it with primes up to the square root of 119, which is approximately 11. (If 119 is not prime, then at least one of its prime factors must be smaller than the square root of 119.) It turns out that 119 is divisible by 7. \(119 = 7 \times 17\). Thus, we now have the full prime factorization of 5,950: \(5950 = 2 \times 5 \times 5 \times 7 \times 17\). Finally, we return to the definition of "length." We are given the example that the length of 20 is 3, since \(2 \times 2 \times 5 = 20\). So we can see that "length" is just the number of prime numbers in the prime factorization, counting repeats (such as 2 in the example of 20). Thus, the length of 5,950 is 5. Answer: E. Check similar questions to practice: foranyintegerk1thetermlengthofaninteger108124.htmlforanypositiveintegernthelengthofnisdefinedas126740.htmlforanypositiveintegernn1thelengthofnisthe126368.htmlthelengthofintegerxreferstothenumberofprime132624.html
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Re: The "length" of a positive integer greater than 1 is the number of pri
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03 Apr 2017, 23:15
5950=595*10= 2*5^2*7*17 5950 has four distinct prime factors BUT 5 is the length of 5950 because we have to consider repeats without writing the number in the exponential form.
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Re: The "length" of a positive integer greater than 1 is the number of pri
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04 Apr 2017, 10:14
Answer is E:
5950/2= 2975 2975/5=595 595/5=123 123/3=31
Therefore 2x5x5x3x31=5950
And the length is 5
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Re: The "length" of a positive integer greater than 1 is the number of pri
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04 Apr 2017, 10:40
Bunuel wrote: Tough and Tricky questions: Number Properties. The "length" of a positive integer greater than 1 is the number of prime numbers, including repeats, that are factors of that integer. For instance, the length of 20 is 3, because \(2 \times 2 \times 5 = 20\). What is the length of 5,950? A. 1 B. 2 C. 3 D. 4 E. 5 Kudos for a correct solution.5,950/50=119 (50)*(119)=(2*5*5)*(7*17) 5 E



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Re: The "length" of a positive integer greater than 1 is the number of pri
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06 Apr 2017, 09:39
Bunuel wrote: Tough and Tricky questions: Number Properties. The "length" of a positive integer greater than 1 is the number of prime numbers, including repeats, that are factors of that integer. For instance, the length of 20 is 3, because \(2 \times 2 \times 5 = 20\). What is the length of 5,950? A. 1 B. 2 C. 3 D. 4 E. 5 Let’s break 5,950 into prime factors. 5,950 = 595 x 10 = 119 x 5 x 5 x 2 = 7 x 17 x 5 x 5 x 2 Thus, the length is 5. Answer: E
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Re: The "length" of a positive integer greater than 1 is the number of pri
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25 May 2018, 04:12
Bunuel wrote: Tough and Tricky questions: Number Properties. The "length" of a positive integer greater than 1 is the number of prime numbers, including repeats, that are factors of that integer. For instance, the length of 20 is 3, because \(2 \times 2 \times 5 = 20\). What is the length of 5,950? A. 1 B. 2 C. 3 D. 4 E. 5 Kudos for a correct solution. 5950 = 2*5*5*7*17
Answer is E = 5
119 tricked me for a while.
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Re: The "length" of a positive integer greater than 1 is the number of pri
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07 Jun 2018, 10:53
5950 = 5*5*2*7*17 all primes (Taking L.C.M )



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Re: The "length" of a positive integer greater than 1 is the number of pri
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12 Jun 2018, 06:41
5950=2*5*5*7*17, the lenght is 5, E



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The "length" of a positive integer greater than 1 is the number of pri
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29 Jun 2018, 08:49
[quote="Bunuel"] Tough and Tricky questions: Number Properties. The "length" of a positive integer greater than 1 is the number of prime numbers, including repeats, that are factors of that integer. For instance, the length of 20 is 3, because \(2 \times 2 \times 5 = 20\). What is the length of 5,950? A. 1 B. 2 C. 3 D. 4 E. 5 5950=2*5*5*7*17 so ans.E
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The "length" of a positive integer greater than 1 is the number of pri &nbs
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