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If m is a positive integer, then m^3 has how many digits?

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If m is a positive integer, then m^3 has how many digits?  [#permalink]

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New post Updated on: 01 Oct 2018, 02:30
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If m is a positive integer, then m^3 has how many digits?

(1) m has 3 digits.
(2) m^2 has 5 digits.


Data Sufficiency
Question: 104
Category: Arithmetic Properties of numbers
Page: 160
Difficulty: 650


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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?  [#permalink]

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New post 20 Feb 2014, 00:17
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31
SOLUTION

If m is a positive integer, then m^3 has how many digits?

Pick some easy numbers.

(1) m has 3 digits --> if \(m=100=10^2\) then \(m^3=10^6\) so it will have 7 digits but if \(m=300=3*10^2\) then \(m^3=27*10^6\) so it will have 8 digits. Not sufficient.

(2) m^2 has 5 digits --> the same values of \(m\) (100 and 300) satisfy this statement too (because if \(m=10^2\) then \(m^2=10^4\) and has 5 digits and if \(m=3*10^2\) then \(m^2=9*10^4\) also has 5 digits), so \(m^3\) may still have 7 or 8 digits. Not sufficient.

(1)+(2) The same example worked for both statements so even taken together statements are not sufficient.

Answer: E.
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Re: If m is a positive integer, then m^3 has how many digits?  [#permalink]

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New post Updated on: 22 Dec 2010, 06:14
13
1
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 \(10^(n-1)\)
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. 12-1)
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. 9-1)
hence NOT suff.

stmnt1&2 together: We can conclude that \(m^3\) can have minimum of 6 and max of 8 digits(i.e. 12-1) ==> m can have 7 or 8 digits
hence NOT suff.

Answer "E".

Regards,
Murali.
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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?  [#permalink]

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New post 22 Dec 2010, 05:21
1
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. 12-1)
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. 9-1)
hence NOT suff.

stmnt1&2 together: We can conclude that \(m^3\) can have minimum of 6 and max of 8 digits(i.e. 12-1) ==> 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?  [#permalink]

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New post 22 Dec 2010, 11:18
8
2
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?  [#permalink]

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New post 21 Feb 2014, 05:39
2
1
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?  [#permalink]

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New post 23 May 2014, 03:02
3
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 9

Statement 1 not sufficient

Statement 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 8

Statement 2 not sufficient

Combining both statement,
Number of digits could be 7 or 8

Therefore,
Answer: E
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Re: If m is a positive integer, then m^3 has how many digits?  [#permalink]

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New post 09 Dec 2017, 16:15
1
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:

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Re: If m is a positive integer, then m^3 has how many digits?  [#permalink]

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