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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] New post 01 Jun 2009, 17:49
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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
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Re: Digits [#permalink] New post 01 Jun 2009, 18:47
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(1) Let M = 100, M^3 = 1000000 (7 digits)
M = 500, M^3 = 125000000 (9 digits)


(2) M = 100, M^2 = 10000 (5 digits), M^3 = 1000000 (7 digits)
M = 300, M^2 = 90000 (5 digits), M^3 = 27000000 (8 digits)

Combining,

Still NOT SUFFICIENT.

Thus, E.
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Re: Digits [#permalink] New post 01 Jun 2009, 21:52
sondenso 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


(1)

X,Y,Z are the 3 digits of M respectively, X being the hundreds, Y the tens and Z the unit.

M = 100X + 10Y + Z

M^{3} = (100X+10Y+Z)^{3}

uh oh... here we go...I did the factoring on paper

M^{3} = 10^{6}(X^{3}) + 10^{5}(3X^{2}Y) + 10^{4}(3X^{2}Z+3XY^{2}) + 10^{3}(6XYZ+y^{3}) + 10^2(3XZ^{2}+3Y^{2}Z) + 10^{1}(3YZ^{2}) + 10^{0}(Z^{3})

10^6 represents the 7th digit-- but if X^{3}>=10, then we really have the 8th digit, as we'd be multiplying 10^6 by 10, which would make it the 8th digit.

So if we look at strictly the digit X, and say it's 1, then X^3 would be 1, and we'd have a 1 in the 7th digit since we're multiplying by 10^6 and putting it into the 7th digit. Now the thing we have to watch out for is in the 6th digit spot, 10^5*(3X^{2}Y), and if X=1 then it's 10^5*(3Y), but 0<=Y<=9, so the most we could have inside there would be 27, which would make it 10^{5}(27) = 10^{6}(2.7)
adding it together we'd get 10^{6}(1) + 10^{6}(2.7 = 10^{6}(3.7), so for sure if X is 1 we'd have 7 or fewer digits. The other digits in the sum are irrelevant, they wouldn't be big enough to affect it.

To put this into perspective,

Say M = 100 (X=1,Y=0,Z=0), then...

M^{3} = 10^{6}(1) + 10^{5}(0) + 10^{4}(0) + 10^{3}(0) + 10^2(0) + 10^{1}(0) + 10^{0}(0)
M^{3} = 10^{6}(1) + 0 + 0 + 0 + 0 + 0 + 0
M^{3} = 10^{6}
M^{3} = 1000000

M has 7 digits.

Say M = 220 (X=2,Y=2,Z=0), then...

M^{3} = 10^{6}(8) + 10^{5}(24) + 10^{4}(24) + 10^{3}(8) + 10^2(0) + 10^{1}(0) + 10^{0}(0)

M^{3} = 10^{6}(10.4) + 10^{5}(2.4) + 10^{3}(8) + 10^2(0) + 10^{1}(0) + 10^{0}(0)

M^{3} = 10^{6}(10 + 4/10) + 10^{5}(2 + 4/10) + 10^{3}(8) + 10^2(0) + 10^{1}(0) + 10^{0}(0)

M^{3} = 10^{6}(10) + 10^{6}(4/10) + 10^{5}(2) + 10^{5}(4/10) + 10^{3}(8) + 10^2(0) + 10^{1}(0) + 10^{0}(0)

M^{3} = 10^{7}(1) + 10^{5}(4) + 10^{5}(2) + 10^{4}(4) + 10^{3}(8) + 10^2(0) + 10^{1}(0) + 10^{0}(0)

M^{3} = 10^{7}(1) + 10^{5}(6) + 10^{4}(4) + 10^{3}(8) + 10^2(0) + 10^{1}(0) + 10^{0}(0)

So M = 10648000

8 digits.

INSUFFICIENT (lol).

(2) M^2 has 5 digits.

This one we can quickly check.

M=90 M^2 = 8100.
M=99 M^2 = 9801
M=100 M^2 = 10000
M=300 M^2 = 90000
M=400 M^2 = 160000

So M is between 100 & 400.

But now looking at the 2 examples from (1), we see that 100^3 has 7 digits 220^3 has 8 digits

INSUFFICIENT

M^2 = 10^{4}(X^{2}) + 10^{3}(2XY) + 10^{2}(2XZ+Y^{2}) + 10^{1}(2YZ) + 10^{0}(Z^{2})

(1&2)

From 2 we have that 100<=M<400

From (2) we have a range for X
1<=X<=3

We can just reuse the 2 examples from (1), since they're still valid when you assume (2) to be true. From the long ass equation from (1), we can see that if M=100, M^3 has 7 digits, but M=220 has 8 digits.

INSUFFICIENT

Final Answer, E.
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Re: Digits [#permalink] New post 02 Jun 2009, 00:05
IMO E, INsufficient even after combining
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Re: Digits [#permalink] New post 02 Jun 2009, 00:09
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Re: Digits [#permalink] New post 02 Jun 2009, 06:33
sondenso 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


(1) If M has 3 digits, M could be 100 or 999.
If M = 100, M^3 = 1,000,000. So 7 digits.
If M = 999, M^3 is close to = 1000,000,000. So 9 digits.

NSF.
(2) If M^2 has 5 digits, M could be 100 or 999.

If M = 100, M^2 = 10,000. So 5 digits.
If M = 999, M^2 is close to = 1,000,000. So 5 digits.

Again M could be 100 or 999. NSF>
Together also same. E.

Hades wrote:
sondenso 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


(1)

X,Y,Z are the 3 digits of M respectively, X being the hundreds, Y the tens and Z the unit.

M = 100X + 10Y + Z

M^{3} = (100X+10Y+Z)^{3}

uh oh... here we go...I did the factoring on paper

M^{3} = 10^{6}(X^{3}) + 10^{5}(3X^{2}Y) + 10^{4}(3X^{2}Z+3XY^{2}) + 10^{3}(6XYZ+y^{3}) + 10^2(3XZ^{2}+3Y^{2}Z) + 10^{1}(3YZ^{2}) + 10^{0}(Z^{3})

10^6 represents the 7th digit-- but if X^{3}>=10, then we really have the 8th digit, as we'd be multiplying 10^6 by 10, which would make it the 8th digit.

So if we look at strictly the digit X, and say it's 1, then X^3 would be 1, and we'd have a 1 in the 7th digit since we're multiplying by 10^6 and putting it into the 7th digit. Now the thing we have to watch out for is in the 6th digit spot, 10^5*(3X^{2}Y), and if X=1 then it's 10^5*(3Y), but 0<=Y<=9, so the most we could have inside there would be 27, which would make it 10^{5}(27) = 10^{6}(2.7)
adding it together we'd get 10^{6}(1) + 10^{6}(2.7 = 10^{6}(3.7), so for sure if X is 1 we'd have 7 or fewer digits. The other digits in the sum are irrelevant, they wouldn't be big enough to affect it.

To put this into perspective,

Say M = 100 (X=1,Y=0,Z=0), then...

M^{3} = 10^{6}(1) + 10^{5}(0) + 10^{4}(0) + 10^{3}(0) + 10^2(0) + 10^{1}(0) + 10^{0}(0)
M^{3} = 10^{6}(1) + 0 + 0 + 0 + 0 + 0 + 0
M^{3} = 10^{6}
M^{3} = 1000000

M has 7 digits.

Say M = 220 (X=2,Y=2,Z=0), then...

M^{3} = 10^{6}(8) + 10^{5}(24) + 10^{4}(24) + 10^{3}(8) + 10^2(0) + 10^{1}(0) + 10^{0}(0)

M^{3} = 10^{6}(10.4) + 10^{5}(2.4) + 10^{3}(8) + 10^2(0) + 10^{1}(0) + 10^{0}(0)

M^{3} = 10^{6}(10 + 4/10) + 10^{5}(2 + 4/10) + 10^{3}(8) + 10^2(0) + 10^{1}(0) + 10^{0}(0)

M^{3} = 10^{6}(10) + 10^{6}(4/10) + 10^{5}(2) + 10^{5}(4/10) + 10^{3}(8) + 10^2(0) + 10^{1}(0) + 10^{0}(0)

M^{3} = 10^{7}(1) + 10^{5}(4) + 10^{5}(2) + 10^{4}(4) + 10^{3}(8) + 10^2(0) + 10^{1}(0) + 10^{0}(0)

M^{3} = 10^{7}(1) + 10^{5}(6) + 10^{4}(4) + 10^{3}(8) + 10^2(0) + 10^{1}(0) + 10^{0}(0)

So M = 10648000

8 digits.

INSUFFICIENT (lol).

(2) M^2 has 5 digits.

This one we can quickly check.

M=90 M^2 = 8100.
M=99 M^2 = 9801
M=100 M^2 = 10000
M=300 M^2 = 90000
M=400 M^2 = 160000

So M is between 100 & 400.

But now looking at the 2 examples from (1), we see that 100^3 has 7 digits 220^3 has 8 digits

INSUFFICIENT

M^2 = 10^{4}(X^{2}) + 10^{3}(2XY) + 10^{2}(2XZ+Y^{2}) + 10^{1}(2YZ) + 10^{0}(Z^{2})

(1&2)

From 2 we have that 100<=M<400

From (2) we have a range for X
1<=X<=3

We can just reuse the 2 examples from (1), since they're still valid when you assume (2) to be true. From the long ass equation from (1), we can see that if M=100, M^3 has 7 digits, but M=220 has 8 digits.

INSUFFICIENT

Final Answer, E.



Thats definitely good effort and nice way to solve the question but is little lengthy and time consuming. GMAT questions such as this one should be solved in around 2 minuets.

I like it. Good job and keep it up. 8-)
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Re: Digits [#permalink] New post 02 Jun 2009, 17:58
GMAT TIGER wrote:
sondenso 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


(1) If M has 3 digits, M could be 100 or 999.
If M = 100, M^3 = 1,000,000. So 7 digits.
If M = 999, M^3 is close to = 1000,000,000. So 9 digits.

NSF.
(2) If M^2 has 5 digits, M could be 100 or 999.

If M = 100, M^2 = 10,000. So 5 digits.
If M = 999, M^2 is close to = 1,000,000. So 5 digits.

Again M could be 100 or 999. NSF>
Together also same. E.

Hades wrote:
sondenso 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


(1)

X,Y,Z are the 3 digits of M respectively, X being the hundreds, Y the tens and Z the unit.

M = 100X + 10Y + Z

M^{3} = (100X+10Y+Z)^{3}

uh oh... here we go...I did the factoring on paper

M^{3} = 10^{6}(X^{3}) + 10^{5}(3X^{2}Y) + 10^{4}(3X^{2}Z+3XY^{2}) + 10^{3}(6XYZ+y^{3}) + 10^2(3XZ^{2}+3Y^{2}Z) + 10^{1}(3YZ^{2}) + 10^{0}(Z^{3})

10^6 represents the 7th digit-- but if X^{3}>=10, then we really have the 8th digit, as we'd be multiplying 10^6 by 10, which would make it the 8th digit.

So if we look at strictly the digit X, and say it's 1, then X^3 would be 1, and we'd have a 1 in the 7th digit since we're multiplying by 10^6 and putting it into the 7th digit. Now the thing we have to watch out for is in the 6th digit spot, 10^5*(3X^{2}Y), and if X=1 then it's 10^5*(3Y), but 0<=Y<=9, so the most we could have inside there would be 27, which would make it 10^{5}(27) = 10^{6}(2.7)
adding it together we'd get 10^{6}(1) + 10^{6}(2.7 = 10^{6}(3.7), so for sure if X is 1 we'd have 7 or fewer digits. The other digits in the sum are irrelevant, they wouldn't be big enough to affect it.

To put this into perspective,

Say M = 100 (X=1,Y=0,Z=0), then...

M^{3} = 10^{6}(1) + 10^{5}(0) + 10^{4}(0) + 10^{3}(0) + 10^2(0) + 10^{1}(0) + 10^{0}(0)
M^{3} = 10^{6}(1) + 0 + 0 + 0 + 0 + 0 + 0
M^{3} = 10^{6}
M^{3} = 1000000

M has 7 digits.

Say M = 220 (X=2,Y=2,Z=0), then...

M^{3} = 10^{6}(8) + 10^{5}(24) + 10^{4}(24) + 10^{3}(8) + 10^2(0) + 10^{1}(0) + 10^{0}(0)

M^{3} = 10^{6}(10.4) + 10^{5}(2.4) + 10^{3}(8) + 10^2(0) + 10^{1}(0) + 10^{0}(0)

M^{3} = 10^{6}(10 + 4/10) + 10^{5}(2 + 4/10) + 10^{3}(8) + 10^2(0) + 10^{1}(0) + 10^{0}(0)

M^{3} = 10^{6}(10) + 10^{6}(4/10) + 10^{5}(2) + 10^{5}(4/10) + 10^{3}(8) + 10^2(0) + 10^{1}(0) + 10^{0}(0)

M^{3} = 10^{7}(1) + 10^{5}(4) + 10^{5}(2) + 10^{4}(4) + 10^{3}(8) + 10^2(0) + 10^{1}(0) + 10^{0}(0)

M^{3} = 10^{7}(1) + 10^{5}(6) + 10^{4}(4) + 10^{3}(8) + 10^2(0) + 10^{1}(0) + 10^{0}(0)

So M = 10648000

8 digits.

INSUFFICIENT (lol).

(2) M^2 has 5 digits.

This one we can quickly check.

M=90 M^2 = 8100.
M=99 M^2 = 9801
M=100 M^2 = 10000
M=300 M^2 = 90000
M=400 M^2 = 160000

So M is between 100 & 400.

But now looking at the 2 examples from (1), we see that 100^3 has 7 digits 220^3 has 8 digits

INSUFFICIENT

M^2 = 10^{4}(X^{2}) + 10^{3}(2XY) + 10^{2}(2XZ+Y^{2}) + 10^{1}(2YZ) + 10^{0}(Z^{2})

(1&2)

From 2 we have that 100<=M<400

From (2) we have a range for X
1<=X<=3

We can just reuse the 2 examples from (1), since they're still valid when you assume (2) to be true. From the long ass equation from (1), we can see that if M=100, M^3 has 7 digits, but M=220 has 8 digits.

INSUFFICIENT

Final Answer, E.



Thats definitely good effort and nice way to solve the question but is little lengthy and time consuming. GMAT questions such as this one should be solved in around 2 minuets.

I like it. Good job and keep it up. 8-)


For the basic guys like me, this approach is very preferable! Thanks Hades
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Re: Digits [#permalink] New post 03 Jun 2009, 22:45
My approach..

1. m is of 3 digits.

cube 100 has 7 digits.
cube 200 has 7 digits but cube 300 has 8 digits.. as 27 adds 2 didits...

2. square M has 5 digits..

square 100 has 5 sigits.. cube 100 has 7 digits..
sqauree 300 has 5 digits.. but cube 300 has 8 digits..

combined 1 and 2, ( actually the second condition already combines the first one..)

still E.
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Re: Digits [#permalink] New post 05 Jun 2009, 12:53
sondenso 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


(1) 100 < = m < 1000
Thus 10^6 <= m < 10^9

m^3 could have 7,8 or 9 digits

Not suff

(2) 10000 < = M^2 < 100000
10^2 < M < = sqrt (10) x 10^2
10^6 < = m^3 < 10sqrt(10) x 10^6

m^3 could have 7 or 8 digits

Not suff

Together not suff
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Re: Digits [#permalink] New post 08 Jun 2009, 23:42
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goldeneagle94 wrote:
(1) Let M = 100, M^3 = 1000000 (7 digits)
M = 500, M^3 = 125000000 (9 digits)


(2) M = 100, M^2 = 10000 (5 digits), M^3 = 1000000 (7 digits)
M = 300, M^2 = 90000 (5 digits), M^3 = 27000000 (8 digits)

Combining,

Still NOT SUFFICIENT.

Thus, E.

In this kind of question, just give example... + 1 Kudo for this post
Re: Digits   [#permalink] 08 Jun 2009, 23:42
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