If t = 1/2^9 * 5^3 is expressed as a terminating decimal, how many zeros will t have between the decimal point and the first nonzero digit to the right of the decimal point? Any theories?
The digits to the right of the decimal point indicate division by a power of 10:diegocuenca wrote:If t = 1/2^9 * 5^3 is expressed as a terminating decimal, how many zeros will t have between the decimal point and the first nonzero digit to the right of the decimal point? Any theories?
.1 = 1/10 = 1/10¹.
.01 = 1/100 = 1/10².
.001 = 1/1000 = 1/10³.
To determine the number of zeros that will appear to the right of the decimal point when 1/(2�5³) is put into decimal notation, factor out as many 10's as possible from the denominator:
2�5³
=2� * (2³5³)
= 64 * (2*5)³
= 64 * 10³
= 64000.
Thus, 1/(2�5³) = 1/64000 = .00001...
There are 4 zeros to the right of the decimal point.












