For any positive integer n, n1, the "length" of n is the

Question

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

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

Bunuel · Accepted Answer

Basically the length of an integer is the sum of the powers of its prime factors. For example the length of 24 is 4 because 24=2^3*3^1 --> 3+1=4. Now, to maximize the length of an integer less then 1,000 we should minimize its prime base(s). Minimum prime base is 2: so 2^x<1,000 --> x<10 --> maximum length is 9 for 2^9=512. Note that 2^9 is not the only integer whose length is 9, for example 2^8*3=768<100 also has the length of 8+1=9. Answer: B. Hope it helps.

Stiv · Answer

To maximize the length you should use the smallest prime number, 2.
2x2x2x2x2x2x2x2x2 = 2^9 = 512; 2^10 = 1024 which is > 1000, so you have to use 2^9.
The answer is B.

Bunuel · Answer

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.

SaraLotfy · Answer

its really helpful to remember here that 2^10 = 1024 , so the second smallest integer less than 1024 would be 2^9

PareshGmat · Answer

To have "Maximum length", base should be least.....

2^{10} = 1024

less than 1000 is  2^9

Answer = B = 9

ScottTargetTestPrep · Answer

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

suganyam · Answer

2^10 = 1024 2^9=512 512 *3 =1536 >1000 henceforth its 9

DanTe02 · Answer

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

CrackverbalGMAT · Answer

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)

D.S
GMAT SME

morhom123 · Answer

Would go with Stiv's solution rather than Bunuel's since 1 is not a prime.

Ze3rebz · Answer

The way I did it was I found the prime factors of 1000 first, which are  2^3 and 5^3 . Now, since 2 is the smallest prime, we will leave 2^3 and focus on  5^3 .

We want to substitute 5^3 by the smallest prime to get the greatest length, which is 2, since  5^3 = 125 . Find the power of 2 that gives a product less than 125, which is 2^6 (2^6 = 64, 2^7 = 128). At last, we are left with  2^3 * 2^6 = 2^9

Answer B

EtaCarnia · Answer

what if the question would have said distinct factors. ? How to solve it then?

For any positive integer n, n>1, the "length" of n is the

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