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705-805 (Hard)|   Fractions and Ratios|                                 
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Runirish
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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muralimba
Runirish
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.
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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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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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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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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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Statement 1-
2≤log m<3
6≤ 3log m<9

Number of digits in \(m^3\) can be 7, 8 or 9
Insufficient

Statement 2-

4≤ 2log m< 5
6≤ 3log m< 7.5

Number of digits in \(m^3\) can be 7 or 8
Insufficient

Combing both equations
Number of digits in \(m^3\) can be 7 or 8

Insufficient

E


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

(1) m has 3 digits.
(2) m^2 has 5 digits.
 
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Bunuel
If m is a positive integer, then m^3 has how many digits?

(1) m has 3 digits.
(2) m^2 has 5 digits.
The best logical approach is to test maximum and minimum values.
Here's how the question can be understood with smaller numbers.

St.1
If m is such that m^3 has digits different(more) than m then the answer would be 'INSUFFICIENT'.
For example: m = 1, m^3 = 1; OR m = 2, m^3 = 8; same number of digits BUT
m = 3, m^3 = 27 - the number of digits are different.

INSUFFICIENT.

St. 2
Similar logic as applied in St. 1 can be applied here. If m = 1, m^2 = 1, m^3 = 1; OR m = 4, m^2 = 16, m^3 = 64 AND
m = 5, m^2 = 25, m^3 = 125 - the number of digits are different.

INSUFFICIENT.

Together St. 1 and St. 2
The logic still prevails even after the two statements are put together.

The examples can be extrapolated and checked for the conditions.

Answer E.
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I think just looking at normal squares and cubes are enough to estimate.

squares 1,4,9,16
cubes 1,8,27,81

cube becomes 2-digit for number 3 while still keeping square to 1-digit.

so 300 is a good candidate.

300^2= 90000 (5 digits)
300^3= 27000000 (8 digits)

compare with lowest 100.
100^2 => 5 digits
100^3 => 7 digits

Ans E
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how can you be sure that 999^3 will have one less digit than 1000^3? it could have any number of digit may be even 10 digits or may be 7 digits or 8 digits.

KarishmaB
KarishmaB


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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GulfTube
how can you be sure that 999^3 will have one less digit than 1000^3? it could have any number of digit may be even 10 digits or may be 7 digits or 8 digits.

KarishmaB


\(1000^3 =1,000,000,000\) which is the smallest 10 digit number. All numbers below it have 9 or fewer digits.
\(900^3 =729,000,000\) is a 9 digit number. All numbers greater have 9 or more digits.

Any number in between the two including \(999^3\) will have 9 digits then.
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Hi GulfTube,

Good instinct to challenge that step. You're reacting to the reasoning (from KarishmaB and rajatjain14) that since 1000^3 = 10^9 is the smallest 10-digit number, 999^3 must have 9 digits. Let me show you why that's locked in and can't wander to 10, 8, or 7.

Why not 10 digits

The key fact: 999 is less than1000, and cubing keeps that order. So

- 999^3 < 1000^3 = 1,000,000,000

And 1,000,000,000 is the smallest10-digit number. If 999^3 is strictly below it, then 999^3 cannot reach 10 digits. That kills the "maybe 10" worry.

Why not 8 or 7 digits

Now the other direction - 999 is almost as big as 1000, so its cube is almost as big as a billion. Actually 999^3 = 997,002,999. That's just a hair under a billion - clearly up in the 9-digit range (between 100,000,000 and 1,000,000,000). It's nowhere near dropping to 8 or 7 digits.

So 999^3 sits snugly at 9 digits: too big to be 8, too small to be 10.

Lock it in with tiny numbers

Watch the same "just below a power of 10" pattern with smaller cases you can check by hand:

- 9 vs 10: 9^3 = 729 (3 digits), 10^3 = 1000 (4 digits) - one fewer.
- 99 vs 100: 99^3 = 970,299 (6 digits), 100^3 = 1,000,000 (7 digits) - one fewer.

Every time, the number just below a power of 10 lands one digit short of it - never two short, never equal. That's exactly why 999^3 has 9 digits, and why Statement (1) still leaves 7, 8, or 9 as options - not sufficient.

Answer: E

GulfTube
how can you be sure that 999^3 will have one less digit than 1000^3? it could have any number of digit may be even 10 digits or may be 7 digits or 8 digits.

KarishmaB

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