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qtrip
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qtrip
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Thanks Bunuel..I was wondering how something like 11 can be represented. But now I understand why 1024 has to be the right answer.
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Thanks Bunuel..I was wondering how something like 11 can be represented. But now I understand why 1024 has to be the right answer.


11 can be expressed as \(p(10,000)=2^0*3^0*5^0*7^0*11^1\).
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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

Theory on Exponents: math-number-theory-88376.html

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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...
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Bunuel
qtrip
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.

Hope this helps.
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The case of 1. p(what)=1?
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The case of 1. p(what)=1?

If n = 0,
p(0) = 2^0 = 1
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