A computer virus destroys computer memory. On the first day, it destroyed half of this memory. On the second day, it destroyed a third of the memory remaining after the first day; on the 3rd day, it destroyed a fourth of the memory remaining after two days, and on the fourth day, it destroyed a fifth of the memory remaining after 3 days. What part of of all the computer memory was left after those four days?
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Think telescoping product. – amd Mar 15 '17 at 01:25
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Or much easier $\frac 45\frac 34\frac 23 *\frac 12$ – fleablood Mar 15 '17 at 01:31
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Day one you have $\frac{1}{2}$ your memory remaining
Day two you have $\frac{1}{2} * \frac{2}{3}$ of your memory remaining.
Day four you have $\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{4}{5} = \frac{1}{5}$ of your memory remaining.
c..
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