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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]  21 Dec 2010, 13:46
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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 12 Oct 2013, 04:26, edited 1 time in total.
Edited the question.
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Re: DS: Number properties [#permalink]  21 Dec 2010, 14:15
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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.

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.

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Re: DS: Number properties [#permalink]  22 Dec 2010, 04:06
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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.

Regards,
Murali.
Kudos?

Last edited by muralimba on 22 Dec 2010, 06:14, edited 1 time in total.
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Re: DS: Number properties [#permalink]  22 Dec 2010, 05:21
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Expert's post
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.

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: DS: Number properties [#permalink]  22 Dec 2010, 11:18
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Expert's post
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.
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Re: If m is a positive integer, then m^3 has how many digits? 1. [#permalink]  11 Oct 2013, 23:07
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Re: DS: Number properties [#permalink]  07 Apr 2014, 06:09
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 $$10^(n-1)$$
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.

Regards,
Murali.
Kudos?

Murali, While you approach is conceptually solid, it is marred by silly errors. For instance, here "==> $$10^2 <= m < 10^4$$" it should be 10^2 <= m < 10^3 and hence m^3 can have no. of digits from 5-9. Another one was already pointed by Bunuel.
Thanks for sharing nonetheless.
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Re: If m is a positive integer, then m^3 has how many digits? [#permalink]  16 May 2015, 00:59
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Re: If m is a positive integer, then m^3 has how many digits?   [#permalink] 16 May 2015, 00:59
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