TomB wrote:
If t = (1) / (2^9 * 5^3) is expressed as a terminating decimal, how many zeros will it have between the decimal point and the fist nonzero digit to the right of the decimal point?
A. Three
B. Four
C. Five
D. Six
E. Nine
bunnel , can you please explain this problem
Given: \(t=\frac{1}{2^9*5^3}\).
Multiply by \(\frac{5^6}{5^6}\) --> \(t=\frac{5^6}{(2^9*5^3)*5^6}=\frac{25*625}{10^9}=\frac{15625}{10^9}=0.000015625\). Hence \(t\) will have 4 zerose between the decimal point and the fist nonzero digit.
Answer: B.
Or another way \(t=\frac{1}{2^9*5^3}=\frac{1}{(2^3*5^3)*2^6}=\frac{1}{10^3*64}=\frac{1}{64000}\).
Now, \(\frac{1}{64,000}\) is greater than \(\frac{1}{100,000}=0.00001\) and less than \(\frac{1}{10,000}=0.0001\), so \(\frac{1}{64,000}\) is something like \(0.0000xxxx\).
Answer: B.
Can someone please explain how is multiplying 5^6 to the denominator (2^9 * 5^3) get 10^9?