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# How many digits does m^3 have, where m is an integer? (1) m

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How many digits does m^3 have, where m is an integer? (1) m [#permalink]  02 Aug 2006, 21:16
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How many digits does m^3 have, where m is an integer?
(1) m has 3 digits
(2) m^2 has 5 digits.

Try this one!
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B

1) m has three digits .
Hence when m = 100 m^3 has 7 digits
when m = 999 m^3 has > 7 digits.
hence not suff.

2) m^2 has 5 digits.

hence m > 100 and < 319

In both cases m^3 will have 7 digits.....
There is a shortcut....... to calculate digits..... just can't recall it now... have to hit the books in the evening.
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B

St1: 100=<m<1000
so 1,000,000<=m^3<1000,000,000
so digits in m^3 may be 7,8,9 : INSUFF

St2: 10,000<=m^2<100,000
so 100<=m<100*SQRT(2) i.e 100<=m<141
so 1,000,000<=m^3< a number less than 8,000,000 (Cube of 200)
so m^3 have 7 digits.:SUFF
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S2.. 10000 <= m^2 < 99999 ?
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freetheking wrote:
S2.. 10000 <= m^2 < 99999 ?

St2 is m^2 is 5 digits.
that means
10,000<=m^2<100,000
OR
10,000<=m^2<=99,999
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ps_dahiya wrote:
freetheking wrote:
S2.. 10000 <= m^2 < 99999 ?

St2 is m^2 is 5 digits.
that means
10,000<=m^2<100,000
OR
10,000<=m^2<=99,999

May bad. I thought 10,000, not 100,000
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OA is E..

Combine S1 and S2.

m=100 3digits
m^2=10,000 5digits
m^3=1,000,000 7digits
m=300 3digits
m^2=90,000 5digits
m^3=27,000,000 8digits
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ps_dahiya wrote:
St2: 10,000<=m^2<100,000
so 100<=m<100*SQRT(2) i.e 100<=m<141
so 1,000,000<=m^3< a number less than 8,000,000 (Cube of 200)
so m^3 have 7 digits.:SUFF

I see my mistake here.
This should be SQRT(10)
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freetheking wrote:
OA is E..

Combine S1 and S2.

m=100 3digits
m^2=10,000 5digits
m^3=1,000,000 7digits
m=300 3digits
m^2=90,000 5digits
m^3=27,000,000 8digits

Perfect E !!! I guess first time caught jaynayak and dahiya on the wrong foot
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sumitsarkar82 wrote:
freetheking wrote:
OA is E..

Combine S1 and S2.

m=100 3digits
m^2=10,000 5digits
m^3=1,000,000 7digits
m=300 3digits
m^2=90,000 5digits
m^3=27,000,000 8digits

Perfect E !!! I guess first time caught jaynayak and dahiya on the wrong foot

If this had been the case then I would have scored 700+ on my first attempt.
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I was wishing it would be (B) until the very last momemt while solving this. I quickly determined that ST1 is not sufficient by plaugging two simple numbers, and, based on experience, a little voice inside of me wanted to cross out the (B) answer. But since it is not the actual exam test, I decided I go ahead and solve it completely.

For the lower bound, m=100, m^3=100^3 has 7 digits.
I approximated the upper as m=315 from sqrt(99,999) (I came pretty close, the actual is m=316.) While doing (315)^3 I was parying it would not exceed 7 digits, but at the end I was wrong: it had 8 digits. So, ST2 is also not sufficient!

So, it must be (E)

That's a very good question, a challenging one for the 2 min time frame.
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tough!!!

Is this the kind of stuff we should expect on the GMAT, or does it get worse? Can someone who cracked a high math score comment?
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Re: DS:digit [#permalink]  05 Aug 2006, 21:52
freetheking wrote:
How many digits does m^3 have, where m is an integer?
(1) m has 3 digits
(2) m^2 has 5 digits.

Try this one!

m={I,+, -, 0}

(1) m=100 --> m^3 = 1000000 (7 digits)
m= 300 --> m^3 = 27000000(8 digits)
BCE

(2) m= 300 --> m^2 = 90000
m^3 = 27000000(8 digits)

m=200 --> m^2 = 40000
m^3 = 8000000=(7 digits)
CE

(1) & (2)
m=200 or 300 giving m^3 as(7digits or 8 digits)

Hence E

Heman
Re: DS:digit   [#permalink] 05 Aug 2006, 21:52
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