pranab01 wrote:
Carcass wrote:
How many zeros will the decimal equivalent of 1/{2^(11) * 5^(7)} + 1/{2^(7) *5^(11)} have after the decimal point prior to the first non-zero digit?
(A) 6
(B) 7
(C) 8
(D) 11
(E) 18
There is graphical representation error, even I read incorrectly i read as 1/2^1 * 15^7 but it is 1/{(2^11) * (5^7)} .Hope it helps
Now comming back to question to make the decimal terminating we have to bring the decimal in the form \(2^n * 5^n\)
From the equation \(\frac{1}{2^{11} * 5^7}+ \frac{1}{2^7 * 5^{11}}\)
We take\(\frac{1}{(2^7 * 5^7)} * [\frac{1}{(2^4)} + \frac{1}{(5^4)}]\)
= \(\frac{1}{10^7}\) \([\frac{1}{16}\) + \(\frac{1}{625}]\)
= \(\frac{1}{10^7}\) \([\frac{641}{10^4}]\)
= \(\frac{641}{10^{11}}\)
= 8 decimal point (since 641 >100)
Fine, the error in the formula completely drove me out way.
Just a question: I am not sure I have understood your final line "= 8 decimal point (since 641 >100)". I would have said that
\(\frac{641}{10^{11}}=164*10^{-11}\), thus we have to move the point by 11 places behind and given that three places are numbers different from zero, there are 8 zeros before six, 11-3 = 8