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The function p(n) on non-negative integer n is defined in [#permalink]
28 Apr 2012, 05:05

3

This post was BOOKMARKED

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A

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D

E

Difficulty:

95% (hard)

Question Stats:

31% (03:01) correct
69% (02:28) wrong based on 138 sessions

The function p(n) on non-negative integer n is defined in the following way: the units digit of n is the exponent of 2 in the prime factorization of p(n), the tens digit is the exponent of 3, and in general, for positive integer k, the digit in the 10^(k–1) th place of n is the exponent on the kth smallest prime (compared to the set of all primes) in the prime factorization of p(n). For instance, p(102) = 20, since 20 = (5^1)(3^0)(2^2). What is the smallest positive integer that is not equal to p(n) for any permissible n?

Re: The function p(n) on non-negative integer n is defined in [#permalink]
28 Apr 2012, 05:18

Expert's post

qtrip wrote:

The function p(n) on non-negative integer n is defined in the following way: the units digit of n is the exponent of 2 in the prime factorization of p(n), the tens digit is the exponent of 3, and in general, for positive integer k, the digit in the 10^(k–1) th place of n is the exponent on the kth smallest prime (compared to the set of all primes) in the prime factorization of p(n). For instance, p(102) = 20, since 20 = (5^1)(3^0)(2^2). What is the smallest positive integer that is not equal to p(n) for any permissible n?

(A) 1 (B) 29 (C) 31 (D) 1,024 (E) 2,310

OA after some discussion.

The function basically transforms the digits of integer n into the power of primes: 2, 3, 5, ...

For example: \(p(9)=2^9\); \(p(49)=2^9*3^4\); \(p(349)=2^9*3^4*5^3\); \(p(6349)=2^9*3^4*5^3*7^4\); ...

The question asks for the leas number that cannot be expressed by the function p(n).

So, the digits of n transform to the power and since single digit cannot be more than 10 then p(n) cannot have the power of 10 or higher.

So, the least number that cannot be expressed by the function p(n) is \(2^{10}=1,024\) (n just cannot have 10 as its digit).

Answer: D.

P.S. If you have the OA you have to indicate it under the spoiler. _________________

Re: The function p(n) on non-negative integer n is defined in [#permalink]
01 Aug 2013, 19:24

Bunuel wrote:

qtrip wrote:

The function p(n) on non-negative integer n is defined in the following way: the units digit of n is the exponent of 2 in the prime factorization of p(n), the tens digit is the exponent of 3, and in general, for positive integer k, the digit in the 10^(k–1) th place of n is the exponent on the kth smallest prime (compared to the set of all primes) in the prime factorization of p(n). For instance, p(102) = 20, since 20 = (5^1)(3^0)(2^2). What is the smallest positive integer that is not equal to p(n) for any permissible n?

(A) 1 (B) 29 (C) 31 (D) 1,024 (E) 2,310

OA after some discussion.

The function basically transforms the digits of integer n into the power of primes: 2, 3, 5, ...

For example: \(p(9)=2^9\); \(p(49)=2^9*3^4\); \(p(349)=2^9*3^4*5^3\); \(p(6349)=2^9*3^4*5^3*7^4\); ...

The question asks for the leas number that cannot be expressed by the function p(n).

So, the digits of n transform to the power and since single digit cannot be more than 10 then p(n) cannot have the power of 10 or higher.

So, the least number that cannot be expressed by the function p(n) is \(2^{10}=1,024\) (n just cannot have 10 as its digit).

Answer: D.

P.S. If you have the OA you have to indicate it under the spoiler.

Hi Bunuel, I am a bit confused here.Cant p(1024) be 2^4 *3^2*5^0*7^1.. kindly elaborate...i get what you mean but am unable to implement it here...

Re: The function p(n) on non-negative integer n is defined in [#permalink]
02 Aug 2013, 00:22

1

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

up4gmat wrote:

Bunuel wrote:

qtrip wrote:

The function p(n) on non-negative integer n is defined in the following way: the units digit of n is the exponent of 2 in the prime factorization of p(n), the tens digit is the exponent of 3, and in general, for positive integer k, the digit in the 10^(k–1) th place of n is the exponent on the kth smallest prime (compared to the set of all primes) in the prime factorization of p(n). For instance, p(102) = 20, since 20 = (5^1)(3^0)(2^2). What is the smallest positive integer that is not equal to p(n) for any permissible n?

(A) 1 (B) 29 (C) 31 (D) 1,024 (E) 2,310

OA after some discussion.

The function basically transforms the digits of integer n into the power of primes: 2, 3, 5, ...

For example: \(p(9)=2^9\); \(p(49)=2^9*3^4\); \(p(349)=2^9*3^4*5^3\); \(p(6349)=2^9*3^4*5^3*7^4\); ...

The question asks for the leas number that cannot be expressed by the function p(n).

So, the digits of n transform to the power and since single digit cannot be more than 10 then p(n) cannot have the power of 10 or higher.

So, the least number that cannot be expressed by the function p(n) is \(2^{10}=1,024\) (n just cannot have 10 as its digit).

Answer: D.

P.S. If you have the OA you have to indicate it under the spoiler.

Hi Bunuel, I am a bit confused here.Cant p(1024) be 2^4 *3^2*5^0*7^1.. kindly elaborate...i get what you mean but am unable to implement it here...

We don't have to find p(1024). In-fact, the question asks to find the value of the smallest integer which can never be assumed by the function p(n), for any non-negative integer,n. For eg, for p(n) =5, the initial integer n = 100, for p(n) = 7, n = 1000 and so on. Now, if p(n) were to be \(1024 = 2^{10}\), that would mean that the units digit of n was 10, which is not possible.

Re: The function p(n) on non-negative integer n is defined in [#permalink]
26 Oct 2014, 20:42

Expert's post

qtrip wrote:

The function p(n) on non-negative integer n is defined in the following way: the units digit of n is the exponent of 2 in the prime factorization of p(n), the tens digit is the exponent of 3, and in general, for positive integer k, the digit in the 10^(k–1) th place of n is the exponent on the kth smallest prime (compared to the set of all primes) in the prime factorization of p(n). For instance, p(102) = 20, since 20 = (5^1)(3^0)(2^2). What is the smallest positive integer that is not equal to p(n) for any permissible n?

(A) 1 (B) 29 (C) 31 (D) 1,024 (E) 2,310

OA after some discussion.

The question asks for the value that p(n) cannot take.

p(n) is of the form \(2^a * 3^b * 5^c * 7^d\)... etc We know that this is prime factorization and that every positive integer can be prime factorized. Then what is the constraint on the value of p(n)? a, b, c, d etc are single digits. So if the prime factorization of a number is \(2^{10}\) or \(3^{24}\), it cannot be p(n).

The smallest value that p(n) cannot take is \(2^{10} = 1024\).

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