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11 Jun 2022, 02:30
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E
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Difficulty:
75%
(hard)
Question Stats:
54% (02:13) correct
46%
(02:12)
wrong
based on 142
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If p is a positive integer, then how many digits p^3 has?
(1) p is a three digit number
(2) p^2 is a five digit number.
Are You Up For the Challenge: 700 Level Questions
(1) p is a three digit number
(2) p^2 is a five digit number.
Are You Up For the Challenge: 700 Level Questions
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11 Jun 2022, 04:38
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If p is a positive integer, then how many digits \(p^3\) has?
(1) \(100 <= p <= 999\)
If \(p = 100, p^3 = 1000000\) (seven digits digits)
If \(p = 900, p^3 = 729000000\) (nine digits).
Insufficient.
Infact, any digit \(>2\) leads to more digits in its cube. \(1^3 = 1. 2^3 = 8. 3^3 = 27\) (one digit increased).
(2) \(10000 <=p^2< 99999\)
\((100)^2 <= p^2 <= (32x)^2\)
So, \(p \) can be anything between 100 and 300 something. We will get a different number of digits in \(p^3\) depending on \(p=100, or, p=300\). So, insufficient.
(1) + (2),
\(p\) can be anything between \(100 \) and \(300 \) something. Insufficient. (E)
(1) \(100 <= p <= 999\)
If \(p = 100, p^3 = 1000000\) (seven digits digits)
If \(p = 900, p^3 = 729000000\) (nine digits).
Insufficient.
Infact, any digit \(>2\) leads to more digits in its cube. \(1^3 = 1. 2^3 = 8. 3^3 = 27\) (one digit increased).
(2) \(10000 <=p^2< 99999\)
\((100)^2 <= p^2 <= (32x)^2\)
So, \(p \) can be anything between 100 and 300 something. We will get a different number of digits in \(p^3\) depending on \(p=100, or, p=300\). So, insufficient.
(1) + (2),
\(p\) can be anything between \(100 \) and \(300 \) something. Insufficient. (E)
29 Jan 2025, 00:46
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No. of digits in \(p^3\)?
Statement 1 - p is 3 digit number
\(100<=p<=999\)
\(10^6 <= p^3 <= ~ 729*10^6\)
Even if we approximate 999 to be 900, then number of digits in \(p^3\) can vary from 7 to 9, hence insufficient.
Statement - \(p^2\) has 5 digits
\(10000 <= p^2 <= 99999\)
\(100 <= p <= ~300\)
\(10^6 <= p^3 <= ~27*10^6\)
Number of digits in \(p^3\) varies from 7 to 8, hence insufficient.
Even combining both the statements, will give number of digits in \(p^3\) to be 7 or 8, hence both together are also insufficient.
IMO: E
Statement 1 - p is 3 digit number
\(100<=p<=999\)
\(10^6 <= p^3 <= ~ 729*10^6\)
Even if we approximate 999 to be 900, then number of digits in \(p^3\) can vary from 7 to 9, hence insufficient.
Statement - \(p^2\) has 5 digits
\(10000 <= p^2 <= 99999\)
\(100 <= p <= ~300\)
\(10^6 <= p^3 <= ~27*10^6\)
Number of digits in \(p^3\) varies from 7 to 8, hence insufficient.
Even combining both the statements, will give number of digits in \(p^3\) to be 7 or 8, hence both together are also insufficient.
IMO: E
