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For any positive integer n, the length of n is defined as th

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For any positive integer n, the length of n is defined as th  [#permalink]

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New post 11 Feb 2010, 16:40
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For any positive integer n, the length of n is defined as the number of prime factors whose product is n. For example, the length of 75 is 3, since 75 = 3 * 5 * 5. How many two-digit positive integers have length 6?

A. None
B. One
C. Two
D. Three
E. Four
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For any positive integer n, the length of n is defined as th  [#permalink]

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New post 29 Jan 2012, 17:21
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Questions about the same concept:

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

Hope it helps.
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Re: 700 Algrbra! Need help again. Thanks so much!  [#permalink]

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New post 11 Feb 2010, 16:55
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For any positive integer n, the length of n is defined as the number of prime factors whose product is n. For example, the length of 75 is 3, since 75 = 3 * 5 * 5. How many two digit positive integers have length 6?

A. None
B. One
C. Two
D. Three
E. Four

Basically the length of the integer is the sum of the powers of its prime factors.

Length of six means that the sum of the powers of primes of the integer (two digit) must be \(6\). First we can conclude that \(5\) can not be a factor of this integer as the smallest integer with the length of six that has \(5\) as prime factor is \(2^5*5=160\) (length=5+1=6), not a two digit integer.

The above means that the primes of the two digit integers we are looking for can be only \(2\) and/or \(3\). \(n=2^p*3^q\), \(p+q=6\) max value of \(p\) and \(q\) is \(6\).

Let's start with the highest value of \(p\):
\(n=2^6*3^0=64\) (length=6+0=6);
\(n=2^5*3^1=96\) (length=5+1=6);

\(n=2^4*3^2=144\) (length=4+2=6) not good as 144 is a three digit integer.

With this approach we see that actually \(5<=p<=6\).

Answer: C.

Hope it helps.
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For any positive integer n, the length of n is defined as number  [#permalink]

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New post 12 Dec 2012, 21:32
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My Solution:

Try increasing prime numbers with length 6:
Trial 1: \(2^6=64\) Valid
Trial 2: \(3^6 =729\) Invalid

This means our candidate 2-digit numbers have combinations of \(2\) and \(3\)

\(2^6=64\)
\(2^5x3^1=96\)
\(2^4x3^2=144\) Invalid

Answer: Two
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Re: 700 Algrbra! Need help again. Thanks so much!  [#permalink]

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New post 11 Feb 2010, 17:23
That's brilliant!!! I especially love the part where I could take 5 away. This really save tons of time! Thanks!! BTW, Thanks so much for the prompt response!
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Re: arithmatic  [#permalink]

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New post 22 Apr 2010, 15:53
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The best (quickest) way I can think of to get the answer, is start with 2^6, then move on from there.

2^6=64
2^5*3=96

Obviously 2^5*5 will be more than 2 digits, as will 2^4*3^2. So 64 and 96 are it. Answer is 2 (C).

You may have been looking for something even faster, but this is fast enough for me. Unless someone has a better way.
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Re: For any positive integer n, the length of n is defined as  [#permalink]

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New post 08 Mar 2013, 16:13
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Try the smallest possible value first: In this case it is 2^6 which equals 64.

If we replace the last 2 with 3, then we have 2^5*3 = 96

From here we can positively assume that any other number will have more than 2 digits. So the answer is (C) 2 numbers that have length 6 and are only 2 digits.
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Re: For any positive integer n, the length of n is defined as  [#permalink]

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New post 01 May 2014, 00:48
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Just writing it out took me .46 sec.:

length of 6, lets take the lowest prime factor 2

2x2x2x2x2x2 = 64

Now substitute the last 2 by a 3, and see the solution gets 96. We can think of what will happen when we substitute another 2 for a three.

Hence, C (2)
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Re: For any positive integer n, the length of n is defined as  [#permalink]

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New post 26 Jul 2016, 19:44
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enigma123 wrote:
For any positive integer n, the length of n is defined as number of prime factors whose product is n, For example, the length of 75 is 3, since 75=3*5*5. How many two-digit positive integers have length 6?

A. 0
B. 1
C. 2
D. 3
E. 4

I need to understand the concept behind solving this question please.

For the length to be 6, the number of prime factors should be maximum. Hence we need to use maximum 2's

The numbers can be 2^6 = 64 and 2^5*3 = 96
For any other number less than 100, the length will be less than 6

Correct option: C
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Re: For any positive integer n, the length of n is defined as  [#permalink]

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New post 22 Mar 2017, 09:36
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enigma123 wrote:
For any positive integer n, the length of n is defined as number of prime factors whose product is n, For example, the length of 75 is 3, since 75=3*5*5. How many two-digit positive integers have length 6?

A. 0
B. 1
C. 2
D. 3
E. 4


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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For any positive integer n, the length of n is defined as  [#permalink]

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New post 26 Mar 2017, 01:26
For any positive integer n, the length of n is defined as number of prime factors whose product is n, For example, the length of 75 is 3, since 75=3*5*5. How many two-digit positive integers have length 6?

A. 0
B. 1
C. 2
D. 3
E. 4

So it is evident that the length can only include 2 and 3 also when we try with 2 we find that length longer than 6 is not possible to be a 2 digit number.

and only one 2 can be replaced by 3 and we get 96,

this can be solved by hit and trial by starting from the lower prime and then moving up.



64 and 96 are the only two numbers possible to have length 6.
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Re: For any positive integer n, the length of n is defined as  [#permalink]

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New post 06 May 2017, 15:40
Bunuel
You made this question look easy. Thanks for your explanation.
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For any positive integer n, the length of n is defined as th  [#permalink]

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New post 29 Aug 2017, 18:21
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YTT wrote:
For any positive integer n, the length of n is defined as the number of prime factors whose product is n. For example, the length of 75 is 3, since 75 = 3 * 5 * 5. How many two-digit positive integers have length 6?

A. None
B. One
C. Two
D. Three
E. Four

Lets start with smallest prime number \(2\).

\(2^6 = 64\) ---------- (Length \(= 6\))

\(2^7\) is three digit number hence cannot be \(n\).

Therefore lets move to next prime number \(3\).

\(2^5*3^1 = 32*3 = 96\) ---------- (Length \(= 6\))

\(2^4*3^2\) will be three digit number, hence cannot be \(n\).

Therefore we have \("Two"\) two-digit positive integers which have length \(6\) \(= 64\) and \(96\)

Answer (C)...
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Re: For any positive integer n, the length of n is defined as th  [#permalink]

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New post 29 Jun 2018, 08:38
YTT wrote:
For any positive integer n, the length of n is defined as the number of prime factors whose product is n. For example, the length of 75 is 3, since 75 = 3 * 5 * 5. How many two-digit positive integers have length 6?

A. None
B. One
C. Two
D. Three
E. Four

2^6=64
2^5*3=96
2^4*3^2=144 out
So ans.C :thumbup:
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Re: For any positive integer n, the length of n is defined as th  [#permalink]

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New post 29 Jul 2018, 09:07
Bunuel wrote:
For any positive integer n, the length of n is defined as the number of prime factors whose product is n. For example, the length of 75 is 3, since 75 = 3 * 5 * 5. How many two digit positive integers have length 6?

A. None
B. One
C. Two
D. Three
E. Four

Basically the length of the integer is the sum of the powers of its prime factors.

Length of six means that the sum of the powers of primes of the integer (two digit) must be \(6\). First we can conclude that \(5\) can not be a factor of this integer as the smallest integer with the length of six that has \(5\) as prime factor is \(2^5*5=160\) (length=5+1=6), not a two digit integer.

The above means that the primes of the two digit integers we are looking for can be only \(2\) and/or \(3\). \(n=2^p*3^q\), \(p+q=6\) max value of \(p\) and \(q\) is \(6\).

Let's start with the highest value of \(p\):
\(n=2^6*3^0=64\) (length=6+0=6);
\(n=2^5*3^1=96\) (length=5+1=6);

\(n=2^4*3^2=144\) (length=4+2=6) not good as 144 is a three digit integer.

With this approach we see that actually \(5<=p<=6\).

Answer: C.

Hope it helps.



Bunuel but \(2^6\) is already 64 and if we multiply it by 3 we get 192

\(n=2^6*3^0=64\) how can it be equal 64 :? (length=6+0=6);
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Re: For any positive integer n, the length of n is defined as th  [#permalink]

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New post 29 Jul 2018, 20:45
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dave13 wrote:
Bunuel wrote:
For any positive integer n, the length of n is defined as the number of prime factors whose product is n. For example, the length of 75 is 3, since 75 = 3 * 5 * 5. How many two digit positive integers have length 6?

A. None
B. One
C. Two
D. Three
E. Four

Basically the length of the integer is the sum of the powers of its prime factors.

Length of six means that the sum of the powers of primes of the integer (two digit) must be \(6\). First we can conclude that \(5\) can not be a factor of this integer as the smallest integer with the length of six that has \(5\) as prime factor is \(2^5*5=160\) (length=5+1=6), not a two digit integer.

The above means that the primes of the two digit integers we are looking for can be only \(2\) and/or \(3\). \(n=2^p*3^q\), \(p+q=6\) max value of \(p\) and \(q\) is \(6\).

Let's start with the highest value of \(p\):
\(n=2^6*3^0=64\) (length=6+0=6);
\(n=2^5*3^1=96\) (length=5+1=6);

\(n=2^4*3^2=144\) (length=4+2=6) not good as 144 is a three digit integer.

With this approach we see that actually \(5<=p<=6\).

Answer: C.

Hope it helps.



Bunuel but \(2^6\) is already 64 and if we multiply it by 3 we get 192

\(n=2^6*3^0=64\) how can it be equal 64 :? (length=6+0=6);


We are not multiplying it by 3, we are multiplying by 3^0, which is 1.
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Re: For any positive integer n, the length of n is defined as th  [#permalink]

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New post 05 Sep 2018, 06:46
We start with taking the least possible two digit number with the length of 6.

In such a case it is,

2*2*2*2*2*2 = 64

We slowly move upwards by increasing the prime factor

2*2*2*2*2*3 = 96

Since taking any other greater prime factor, (5) our number will become 160 which is a three digit number, we stop at 3.

Hence, the answer is TWO

C
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Re: For any positive integer n, the length of n is defined as th   [#permalink] 05 Sep 2018, 06:46
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