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Re: For any positive integer n, n>1, the "length" of n is the [#permalink]
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Splendidgirl666 wrote:
For any positive integer n, n>1, the "length" of n is the number of positive primes (not necessary distinct) whose product is n. For ex, the length of 50 is 3, since 50=2x5x5. What is the greatest possible length of a positive integer less than 1000.

A. 10
B. 9
C. 8
D. 7
E. 6


Questions about the same concept (the length of an integer):

https://gmatclub.com/forum/for-any-posi ... 90320.html
https://gmatclub.com/forum/for-any-posi ... 26368.html
https://gmatclub.com/forum/the-length-o ... 88734.html
https://gmatclub.com/forum/the-length-o ... 32624.html
https://gmatclub.com/forum/for-any-posi ... 40950.html
https://gmatclub.com/forum/for-any-inte ... 08124.html
https://gmatclub.com/forum/what-is-the- ... 32111.html

Hope it helps.
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Re: For any positive integer n, n>1, the "length" of n is the [#permalink]
its really helpful to remember here that 2^10 = 1024 , so the second smallest integer less than 1024 would be 2^9
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Re: For any positive integer n, n>1, the "length" of n is the [#permalink]
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To have "Maximum length", base should be least.....

\(2^{10} = 1024\)

less than 1000 is \(2^9\)

Answer = B = 9
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Re: For any positive integer n, n>1, the "length" of n is the [#permalink]
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Splendidgirl666 wrote:
For any positive integer n, n>1, the "length" of n is the number of positive primes (not necessary distinct) whose product is n. For ex, the length of 50 is 3, since 50=2x5x5. What is the greatest possible length of a positive integer less than 1000.

A. 10
B. 9
C. 8
D. 7
E. 6


In order to maximize the “length,” we need to minimize the values of the prime factors of n. Since 2 is the smallest prime, let’s see how many factors of 2 we can use to get a product less than 1000. Since 2^9 = 512, we see that the largest possible length of a positive integer less than 1000 is 9.

Answer: B
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Re: For any positive integer n, n>1, the "length" of n is the [#permalink]
2^10 = 1024 2^9=512 512 *3 =1536 >1000 henceforth its 9
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Re: For any positive integer n, n>1, the "length" of n is the [#permalink]
Here's an interesting way to do this. No one has shown this way so Ill try
okay so length for less than 1000 we know that 1000= 5^3*2^3 and we want the length to be max so it should be all 2^k form and less than 1000 . Well 4=2^2 is less than 5 so 2^9 will definitely less than 1000
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Re: For any positive integer n, n>1, the "length" of n is the [#permalink]
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To maximize on the length of an integer less then 1,000 we should minimize its prime factor.

The minimum value of such a prime factor is 2 and hence think about the highest power of 2 that is less than 1000.

2^9 = 512

Since the the "length" of n is the number of positive primes (not necessary distinct) whose product is n, length here is 9.

(option b)

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Re: For any positive integer n, n>1, the "length" of n is the [#permalink]
Would go with Stiv's solution rather than Bunuel's since 1 is not a prime.
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Re: For any positive integer n, n>1, the "length" of n is the [#permalink]
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