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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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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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01 Jun 2009, 18:47
1
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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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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

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02 Jun 2009, 00:05
IMO E, INsufficient even after combining
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02 Jun 2009, 00:09
Thanks guys, OA is E
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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.

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

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.
_________________

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GT

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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.

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

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.

For the basic guys like me, this approach is very preferable! Thanks Hades
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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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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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08 Jun 2009, 23:42
1
KUDOS
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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