Data Sufficiency

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

Ans should be B

Stmt 1 Pick nos 100 and 999
100^3 has 7 digits
999^3 has 10 digits

Stmt 2
M^2 has 5 digits => M^3 will have 7 digits

Answer should be E.
1. take the smallest & biggest 3 digit numbers, M can be between: 100 & 999 respectively.
The cubes will have 7 & 9 digits respectively. NOT SUFFICIENT.

2. Locate the smalled & biggest 5 digit numbers: 10000 & 99999 respectively. The roots of these two are: 100 & 316 (approx.) respectively.
So 100 & 316 are the smallest & biggest numbers to have square with 5 digits. (the possible values for M)
For getting the number of digits in M^3
Take a cube of 100: 7 digits
Cube of 316: 8 digits.
Hence again NOT SUFFICIENT.

I agree with Bharat the ans should be E,

however I would explain stmt 2 as follows

M ^ 2 is 5 digit
hence M ^ 2 can be between 10000 & 99999

hence lower limit of M is 100, Upper limit is root of 99999

now since we need to find only the nos of digits & NOT M^3 hence we dont need to calculate root of 99999 as it would consume some time.

we can just say that Lower limit of M is 100 and upper limit is
little above 300 (Closer to 300 i.e 300 ^ 2 =90000)

now 100 ^ 3 =1000000 has 7 digts

now for upper limit we must use M ^ 2 * M

but here we dont know M hence we can say

it is little greater than M ^ 2 * 300

i.e 99999 * 300
i.e (100000 -1 ) *300
i.e 30000000 -300

now this figure has 8 digits so our required will be greater than this by relatively small margin hence it too will have 8 digits

so NOT SUFF.

Good One Samir. I got my mistake. thx

Thanks for the shortcut Samir. Its really faster than the method I used.

correct thanks