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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.

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 _________________

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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