# How to estimate a simple sum on a calculator paper?

The Government has decided to change the GCSE System by adding a 3rd calculator paper and introducing a grading system of $9-1$ ($9$ for those A** kids given to the top $2.5%$ of the country).

I am currently doing some practice questions on this 3rd paper and am struggling with this question. I understand how to estimate however I am not entirely sure how to do this as most of the estimate questions have been on the non-calculator paper which involved rounding to $1$ s.f.

The question:

A virus on a computer is causing errors. An antivirus program is run to remove these errors. An estimate for the number of errors at the end of $t$ hours is $10^6\times2^{-t}$.

Part a) Work out an estimate for the number of errors on the computer at the end of $8$ hours. Worth $2$ marks.

What I did: The first thing that I did was substituted $t$ with $8$ and put it into the calculator getting the answer of $3926.25$ but then I thought... That can't be estimate.

Part b) Explain whether the number of errors on this computer ever reaches zero. (Worth $1$ mark).

I am not sure on part $b$. Should I make the sum equal to $0$ and work from there? The marks are minimal so it must be easier than I think.

For part a, you want an estimate of how many errors are left after 8 hours. By definition, this is given by the formula $10^6 \cdot 2^{-t} = 10^6 \cdot 2^{-8} = 10^6/257 = 3906.25.$ This is your estimate.
Part b cannot be answered, since you only have an estimate of the number of errors. However, looking at the estimate, you notice that $10^6 \cdot 2^{-t}$ is never equal to $0$. So your estimate of number of errors will never equal $0$. But, it will get very close to $0$ as you increase $t$. For example after $t= 24$ hours, the estimated number of errors are down to 0.0596. So while the number of actual errors may or may not reach $0$, the number of estimated errors will never reach $0$, but they will get closer and closer to it as time goes by.