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If m is a positive integer, then m^3 has how many digits?
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Updated on: 01 Oct 2018, 02:30
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53% (01:47) correct 47% (01:37) wrong based on 919 sessions
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Originally posted by Bunuel on 21 Dec 2010, 13:46.
Last edited by Bunuel on 01 Oct 2018, 02:30, edited 2 times in total.
Edited the question.




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If m is a positive integer, then m^3 has how many digits?
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20 Feb 2014, 00:17




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Re: If m is a positive integer, then m^3 has how many digits?
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Updated on: 22 Dec 2010, 06:14
Runirish wrote: If m is a positive integer, then m^3 has how many digits? 1. m has 3 digits 2. m^2 has 5 digits
How would you do this quickly? Is there a rule that I am unaware of? I could do it, but I had to pick a few numbers. Thanks. as we know, the minimum value of a 3 digit integer is 100 = \(10^2\) the maximum value of a 3 digit integer is 999 = \(10^4  1\) the minimum value of a 5 digit integer is 10000 = \(10^4\) the maximum value of a 5 digit integer is 99999 = \(10^6  1\) . . hence, . the minimum value of a \(n\) digit integer is \(10^(n1)\) the maximum value of a \(n\) digit integer is \(10^(n+1)  1\) Back to original qtn: If m is a positive integer, then \(m^3\) has how many digits? stmnt1: \(m\) has 3 digits ==> \(10^2 <= m < 10^4\) ==> \(10^6 <= m^3 < 10^12\) ==> \(m^3\) can have minimum of 7 (i.e 6+1) and max of 11 digits (i.e. 121) hence NOT suff. stmnt2: m^2 has 5 digits ==> \(10^4 <= m^2 < 10^6\) ==> \(10^2 <= m < 10^3\) ==> \(10^6 <= m^3 < 10^9\) ==> \(m^3\) can have minimum of 7 (i.e 6+1) and max of 8 digits (i.e. 91) hence NOT suff. stmnt1&2 together: We can conclude that \(m^3\) can have minimum of 6 and max of 8 digits(i.e. 121) ==> m can have 7 or 8 digits hence NOT suff. Answer "E". Regards, Murali. Kudos?
Originally posted by muralimba on 22 Dec 2010, 04:06.
Last edited by muralimba on 22 Dec 2010, 06:14, edited 1 time in total.




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Re: If m is a positive integer, then m^3 has how many digits?
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22 Dec 2010, 05:21
muralimba wrote: Runirish wrote: If m is a positive integer, then m^3 has how many digits? 1. m has 3 digits 2. m^2 has 5 digits
How would you do this quickly? Is there a rule that I am unaware of? I could do it, but I had to pick a few numbers. Thanks. as we know, the minimum value of a 3 digit integer is 100 = \(10^2\) the maximum value of a 3 digit integer is 999 = \(10^4  1\) the minimum value of a 5 digit integer is 10000 = \(10^4\) the maximum value of a 5 digit integer is 99999 = \(10^6  1\) . . hence, . the minimum value of a \(n\) digit integer is \(10^n\) the maximum value of a \(n\) digit integer is \(10^(n+1)  1\)
Back to original qtn: If m is a positive integer, then \(m^3\) has how many digits? stmnt1: \(m\) has 3 digits ==> \(10^2 <= m < 10^4\) ==> \(10^6 <= m^3 < 10^12\) ==> \(m^3\) can have minimum of 6 and max of 11 digits (i.e. 121) hence NOT suff. stmnt2: m^2 has 5 digits ==> \(10^4 <= m^2 < 10^6\) ==> \(10^2 <= m < 10^3\) ==> \(10^6 <= m^3 < 10^9\) ==> \(m^3\) can have minimum of 6 and max of 8 digits (i.e. 91) hence NOT suff. stmnt1&2 together: We can conclude that \(m^3\) can have minimum of 6 and max of 8 digits(i.e. 121) ==> m can have 6,7, or 8 digits hence NOT suff. Answer "E". Regards, Murali. Kudos? m^3 can have only 7 or 8 digits, not 6. If m=100=10^2 then m^3=10^6 and it has 6 trailing zeros but 7 digits.
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Re: If m is a positive integer, then m^3 has how many digits?
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22 Dec 2010, 11:18
Runirish wrote: If m is a positive integer, then m^3 has how many digits? 1. m has 3 digits 2. m^2 has 5 digits
How would you do this quickly? Is there a rule that I am unaware of? I could do it, but I had to pick a few numbers. Thanks. 1. m has 3 digits When I look at such statements, I invariably think of the extremities. (as muralimba did above) Smallest m = 100 which implies m^3 = 10^6 giving 7 digits. Largest m = 999 but it is not easy to find its cube so I take a number close to it i.e. 1000 and find its cube which is 10^9 i.e. smallest 10 digit number. Hence 999^3 will have 9 digits. Since we can have 7, 8 or 9 digits, this statement is not sufficient. 2. m^2 has 5 digits Now try to forget what you read above. Just focus on this statement. Smallest m^2 = 10000 which implies m = 100 Largest m^2 is less than 99999 which gives m as something above 300 but less than 400. Now, if m is 100, m^3 = 10^6 giving 7 digits. If m is 300, m^3 = 27000000 giving 8 digits. Since we have 7 or 8 digits for m, this statement is not sufficient. Now combining both, remember one important point  If one statement is already included in the other, and the more informative statement is not sufficient alone, both statements will definitely not be sufficient together.e.g. statement 1 tells us that m has 3 digits. Statement 2 tells us that m is between 100 and 300 something, so statement 2 tells us that m has 3 digits (what statement 1 told us) and something extra (that its value lies between 100 and 300 something). Statement 2 is more informative and is not sufficient alone. Since statement 1 doesn't add any new information to statement 2, no way will they both together be sufficient. Hence answer (E).
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Re: If m is a positive integer, then m^3 has how many digits?
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21 Feb 2014, 05:39
Number Picking Strategy:
1)m can have three digits but m^3 differ e.g 100 and 999.Insufficient 2) m^2 can have 5 digits but m^3 differ e.g 100 and 315. Insufficient Together, the statements are insufficient. Take 100 and 315 as in the (2) above.
Answer E.



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Re: If m is a positive integer, then m^3 has how many digits?
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23 May 2014, 03:02
Hi MensaNumber, I hope this helps, Statement 1:M has 3 digits M can be any number between 100 and 999, Number of digits in \(100^3\) = 7 Number of digits in \(1000^3\) = 10 => Hence \(999^3\) will have 9 digits So, number of digits could be 7, 8 or 9Statement 1 not sufficientStatement 2:\(M^2\) has 5 digits Smallest M is 100, where\(M^2\) has 5 digits Largest M would be close to 300, as \(300^2\) = 90000 M=100, will give number of digits in \(M^3\) as 7 M = 300, will give number of digits in \(M^3\) as 8 So, number of digits could be 7 or 8Statement 2 not sufficientCombining both statement, Number of digits could be 7 or 8Therefore, Answer: E
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Re: If m is a positive integer, then m^3 has how many digits?
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09 Dec 2017, 16:15
Hi All, We're told that M is a positive integer. We're asked for the number of digits that M^3 has. This question can be solved by TESTing VALUES. 1) M has 3 digits. IF.... M=100, then M^3 = 1,000,000 and the answer to the question is 7 M=300, then M^3 = 27,000,000 and the answer to the question is 8 Fact 1 is INSUFFICIENT 2) M^2 has 5 digits The same values that we used in Fact 1 also 'fit' Fact 2: M=100, M^2=10,000 and M^3 = 1,000,000 and the answer to the question is 7 M=300, M^2=90,000 and M^3 = 27,000,000 and the answer to the question is 8 Fact 2 is INSUFFICIENT Combined, there's no more work needed. We already have two values that 'fit' both Facts and produce two different answers. Combined, INSUFFICIENT Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: If m is a positive integer, then m^3 has how many digits?
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