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Prep Test Question Again

by tanvis1120 » Wed Aug 20, 2014 6:37 pm
Hi,
Is there any quicker method to solve the below mentioned problem quickly?
Mechanical method may introduce error..
Two numbers are 96 and 64..

Thanks in Advance!
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by GMATGuruNY » Wed Aug 20, 2014 7:17 pm
For any positive integer n the length is defined as the number of prime numbers whose product equals n. So for 75 the length is 3 since 75 = 3 * 5 * 5. How many 2-digit numbers have a length of 6?

a) None
b) One
c) Two
d) Three
e) Four
Start with the smallest possible prime factors:
2*2*2*2*2*2 = 64.
2*2*2*2*2*3 = 96.
Since the product must be less than 100, only the two numbers above are possible. If we increase any of the factors, the product will be greater than 100.

The correct answer is C.
Last edited by GMATGuruNY on Tue Jul 26, 2016 3:10 am, edited 1 time in total.
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by GMATinsight » Wed Aug 20, 2014 8:53 pm
tanvis1120 wrote:Hi,

Thanks in Advance!
Hi Tanvis,

This question doesn't require any method other than Mitch's explanation. For you reference other good question on length of integer is right here...

For any integer k > 1, the term "length of an integer" refers to the number of positive prime factors, not necessarily distinct, whose product is equal to k. For example, if k = 24, the length of k is equal to 4, since 24 = 2 X 2 X 2 X 3. If x and y are positive integers such that x > 1, y > 1, and x + 3y < 1000, what is the maximum possible sum of the length of x and the length of y?

5
6
15
16
18

Answer for This question: Option D
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by Brent@GMATPrepNow » Wed Aug 20, 2014 8:56 pm
For any positive integer n, the length n is defined as the number of prime factors whose product is n. For example, the length of 75 is 3, since 75= 3 X 5 X 5. How many two-digit positive integers have length 6?

A) None
B) One
C) Two
D) Three
E) Four
Let's first find the smallest value with length 6.
This is the case when each prime factor is 2.
We get 2x2x2x2x2x2 = 64. This is a 2-digit positive integer. PERFECT

To find the next largest number with length 6, we'll replace one 2 with a 3
We get 3x2x2x2x2x2 = 96. This is a 2-digit positive integer. PERFECT

To find the third largest number with length 6, we'll replace another 2 with a 3
We get 3x3x2x2x2x2 = 144. This is a 3-digit positive integer. NO GOOD

So there are only two, two-digit positive integers with length 6.

Answer = C

Cheers,
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by [email protected] » Wed Aug 20, 2014 9:06 pm
Hi tanvis1120,

When deciding how to approach a GMAT question, it's important to take advantage of ALL of the information presented. Here, you'll notice that the answer choices are SMALL, meaning that there can only be 0 to 4 possible numbers that fit the description given in the prompt. We're told to multiply 6 prime numbers together and get a total that is only 2 digits (less than 100). There just can't be that many ways for that to happen (and we know there's not, from the answer choices). From here, rather than trying to randomly find numbers that fit, we have to work in reverse and think about what multiplying primes together would get us. The other explanations show you the specific numbers, so I won't rehash that here.

The take-away from this prompt is that you should keep your thinking flexible. Be ready to come up with ways to deal with the problem that aren't what you learned in "math class." That way of thinking is essential to scoring at a high level on the GMAT.

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by Matt@VeritasPrep » Wed Aug 20, 2014 11:21 pm
Tanvis, if you want a good follow up question that explores the concept a little more deeply, try this one:

Let x be the three digit positive integer with the greatest number of factors. If x has n factors, what is the value of n?

A:: 28
B:: 30
C:: 32
D:: 35
E:: 36
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by tanvis1120 » Mon Aug 25, 2014 7:52 pm
Matt@VeritasPrep wrote:Let x be the three digit positive integer with the greatest number of factors. If x has n factors, what is the value of n?

A:: 28
B:: 30
C:: 32
D:: 35
E:: 36
It is 36.
Out of 1,2,3,4,5,6,7,8,9
I multiplied 1,2,3,4,5,6 that gave the three digit number 720. 720 is divisible by all the numbers from 1 to 10 except 7.
It should, therefore, have the highest possibility of the total number of factors.
Factors: 1,2,3,4,5,6,8,9,10, This can be further represented as:
1x1, 1x2, ... 1x10.
In this way, we can form a combination of 9C2 = 36.
Thus, the total number of factors (n) is 36.

Is it right????
@Matt: Please reply

Thanks !
Last edited by tanvis1120 on Mon Aug 25, 2014 8:06 pm, edited 1 time in total.
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by Brent@GMATPrepNow » Mon Aug 25, 2014 7:57 pm
tanvis1120 wrote:
Matt@VeritasPrep wrote:Let x be the three digit positive integer with the greatest number of factors. If x has n factors, what is the value of n?

A:: 28
B:: 30
C:: 32
D:: 35
E:: 36
I could not solve this. It is huge. Is it a real GMAT question??
I considered 6^3 (Of the numbers from 1-9, 6 only has 2 distinct prime factors. I can use combination of its factors 2,3, 6 and the relevant multiples, as the factors of the 3 digit number do not need to be prime).
6^3 = 216
216*4 = 864..
I don't know if it is correct or not.
How can I find all the factors?
If N = (p^a)(q^b)(r^c)..., where p, q, r,...(etc.) are prime numbers, then the total number of positive divisors of N is equal to (a+1)(b+1)(c+1)...

Example: 14000 = (2^4)(5^3)(7^1)
So, the number of positive divisors of 14000 = (4+1)(3+1)(1+1) = 5x4x2=40

Using your example, 864 = (4)(6³)
= (2²)(2³)(3³)
= (2�)(3³)

So, the number of positive divisors of 965 = (5+1)(3+1)
= (6)(4)
= 24

Cheers,
Brent
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by tanvis1120 » Mon Aug 25, 2014 8:07 pm
Brent@GMATPrepNow wrote:
tanvis1120 wrote:
Matt@VeritasPrep wrote:Let x be the three digit positive integer with the greatest number of factors. If x has n factors, what is the value of n?

A:: 28
B:: 30
C:: 32
D:: 35
E:: 36
I could not solve this. It is huge. Is it a real GMAT question??
I considered 6^3 (Of the numbers from 1-9, 6 only has 2 distinct prime factors. I can use combination of its factors 2,3, 6 and the relevant multiples, as the factors of the 3 digit number do not need to be prime).
6^3 = 216
216*4 = 864..
I don't know if it is correct or not.
How can I find all the factors?
If N = (p^a)(q^b)(r^c)..., where p, q, r,...(etc.) are prime numbers, then the total number of positive divisors of N is equal to (a+1)(b+1)(c+1)...

Example: 14000 = (2^4)(5^3)(7^1)
So, the number of positive divisors of 14000 = (4+1)(3+1)(1+1) = 5x4x2=40

Using your example, 864 = (4)(6³)
= (2²)(2³)(3³)
= (2�)(3³)

So, the number of positive divisors of 965 = (5+1)(3+1)
= (6)(4)
= 24

Cheers,
Brent
Oh, Brent, I just solved it above..
Isn't that right too.. ?? :(
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by tanvis1120 » Tue Aug 26, 2014 9:27 am
I did not get any definite answer to this question.
If the three digit number is 720, it can be factorized as = (2^4)(3^2)(5^1)
Therefore, the total number of factors = (4+1) (2+1) (1+1) = 30

What is the answer?
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by Jeff@TargetTestPrep » Wed Dec 13, 2017 9:22 am
For any positive integer n the length is defined as the number of prime numbers whose product equals n. So for 75 the length is 3 since 75 = 3 * 5 * 5. How many 2-digit numbers have a length of 6?

a) None
b) One
c) Two
d) Three
e) Four
We need to determine how many 2-digit integers have a length of 6, or in other words how many 2-digit integers are made up of 6 prime factors. Let's start with the smallest possible numbers:

2^6 = 64 (has a length of 6)

2^5 x 3^1 = 96 (has a length of 6)

Since 2^4 x 3^2 = 144 and 2^5 x 5^1 = 160 are greater than 99, there are no more 2-digit numbers that have a length of 6.

Answer: C

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