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


Let's first find the smallest value with length 6.
This is the case when each prime factor is 2.
We get 2 x 2 x 2 x 2 x 2 x 2 = 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 3 x 2 x 2 x 2 x 2 x 2 = 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 3 x 3 x 2 x 2 x 2 x 2 = 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,
Brent

Originally posted by BrentGMATPrepNow on 07 Sep 2020, 08:09.
Last edited by BrentGMATPrepNow on 10 Jan 2022, 07:25, edited 1 time in total.
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Re: For any positive integer n, the length of n is defined as th [#permalink]
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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?

According to the question,
n=\( a^p*b^q*c^r\)
Length = p+q+r

We need to find two-digit positive integers whose length=6; means, p+q=6

Case 1) \(2^6\); length=6; n=64
Case 2) \(2^5 * 3^1\); length=6; n=96
Case 3) \(2^4 * 3^2\); length=6; n=144 Three-Digit Incorrect
Case 5) \(2^3 * 3^3\); length=6; n=216 Three- Digit Incorrect
As we increase, the value of n will increase beyond 2-digit positive integer


A. None
B. One
C. Two Answer
D. Three
E. Four
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Re: For any positive integer n, the length of n is defined as th [#permalink]
GMATNinja, I determined that the following options are possible:
2^5, ---> (5+1)=6
2^2*3^1--> (2+1)(1+1)=6
2*3^2--> (1+1)*(2+1)=6

Hence, there are 3 possibilities in total. Could you please explain where I am getting wrong with my analysis?
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