m > 0 How many digits m^3 had?

St1: M has 3 digits.
if M = 100, M^3 = 1,000,000 (7 digits)
if M = 900, M^3 = 729,000,000 (9 digits)
Not sufficient

St 2: m^2 has 5 digits
if m = 150, m^2 = 22,500 (5 digits), m^3 = 3,375,000 (7 digits)
if m = 300, m^2 = 90,000 (5 digits), m^3 = 27,000,000 (8 digits)
hence Not Sufficient

Together also Not sufficient (see calc for 150 and 300 in Step 2)

hence E

If M = 100, M^3 = 10,00,000 7 digits
If M = 900, M^3 = 72,90,00,000 9 digits

So, can't say anything.

If M = 100, M^2 = 10,000 (5 digits) and M^3 = 10,00,000 (7 digits)
If M = 300, M^2 = 90,000 (5 digits) and M^3 = 2,70,00,000 (8 digits)

Hence, both statements are insufficient.

Great stuff from the group and I'll just add that on this one your goals should be:

1) Pick numbers (like these guys did) that prove that the information is insufficient. This forces you to think strategically about which numbers to pick, which should lead you to:

2) Look for limits. 100 is the smallest 3-digit number and it has 5 digits when squared and 7 when cubed. So you already have the answer "7" - now you want something bigger. How big can you (easily) go to get a 3-digit number that has 5 when squared? Think of numbers that square to a single digit (like 3^2 = 9) and make that your first digit. 300^2 is just (3*100)^2, and you've already "done the work" for 100^2, so you now have 90000 for 5 digits. But that's about as high as you can go with 5 digits so this is a good one to test for the cube, and you'll see that 3^3 = 27, meaning that you pick up an extra digit off of the first number (3) and the rest (it's (3*100)^3) will stay the same as 100^3, so you now have one more digit than you did before, solidifying that it's not sufficient information.