Statement 1: 10x is not an integerIf x² < x and x is written as a terminating decimal, does x have a nonzero hundredths digit?
(1) 10x is not an integer.
(2) 100x is an integer.
In other words, when the decimal point is moved ONE PLACE TO THE RIGHT, the result is a NON-INTEGER.
It's possible that x = .01, since moving the decimal point one place to the right yields non-integer 0.1.
In this case, the hundredths digit of x is nonzero.
It's possible that x = .001, since moving the decimal point one place to the right yields non-integer 0.01.
In this case, the hundredths digit of x is 0.
INSUFFICIENT.
Statement 2: 100x is an integer
In other words, when the decimal point is moved TWO PLACES TO THE RIGHT, the result is an INTEGER.
It's possible that x = .01, since moving the decimal point two places to the right yields 1.
In this case, x has a nonzero hundredths digit.
It's possible that x = .1, since moving the decimal point two places to the right yields 10.
In this case, x does not have a nonzero hundredths digit.
INSUFFICIENT.
Statements combined:
Since moving the decimal point one place to the right yields a NONINTEGER, while moving the decimal point two places to the right yields an INTEGER, x must be of the form 0.AB, where B is a NONZERO digit.
To illustrate:
0.01 --- one place to the right --> 0.1 --- two places to the right --> 1.
0.32 --- one place to the right --> 3.2 --- two places to the right --> 32.
0.15 --- one place to the right --> 1.5 --- two places to the right --> 15.
If the hundreds digit of x is 0, then both statements cannot be satisfied.
To illustrate:
0.001 --- one place to the right --> 0.01 --- two places to the right --> 0.1.
Since moving two places to the right yields a NON-INTEGER, statement 2 is not satisfied.
Thus, to satisfy both statements, x must have a nonzero hundredths digit.
SUFFICIENT.
The correct answer is C.
