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E
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Question Stats:
56% (02:16) correct
44%
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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.
(1) m has 3 digits.
(2) m^2 has 5 digits.
20 Feb 2014, 01:17
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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.
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.
Runirish
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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