Hi GulfTube,Good instinct to challenge that step. You're reacting to the reasoning (from KarishmaB and rajatjain14) that since
1000^3 =
10^9 is the smallest
10-digit number,
999^3 must have
9 digits. Let me show you why that's locked in and can't wander to
10,
8, or
7.
Why not 10 digitsThe key fact:
999 is
less than1000, and cubing keeps that order. So
-
999^3 <
1000^3 =
1,000,000,000And
1,000,000,000 is the
smallest10-digit number. If
999^3 is strictly below it, then
999^3
cannot reach
10 digits. That kills the "maybe 10" worry.
Why not 8 or 7 digitsNow the other direction -
999 is almost as big as
1000, so its cube is almost as big as a billion. Actually
999^3 =
997,002,999. That's just a hair under a billion - clearly up in the
9-digit range (between
100,000,000 and
1,000,000,000). It's nowhere near dropping to
8 or
7 digits.
So
999^3 sits snugly at
9 digits: too big to be
8, too small to be
10.
Lock it in with tiny numbersWatch the same "just below a power of 10" pattern with smaller cases you can check by hand:
-
9 vs
10:
9^3 =
729 (
3 digits),
10^3 =
1000 (
4 digits) - one fewer.
-
99 vs
100:
99^3 =
970,299 (
6 digits),
100^3 =
1,000,000 (
7 digits) - one fewer.
Every time, the number just below a power of
10 lands one digit short of it - never two short, never equal. That's exactly why
999^3 has
9 digits, and why Statement (
1) still leaves
7,
8, or
9 as options -
not sufficient.
Answer: EGulfTube
how can you be sure that 999^3 will have one less digit than 1000^3? it could have any number of digit may be even 10 digits or may be 7 digits or 8 digits.
KarishmaB