# using trial and error in math problems

Suppose one bacterium is in a jar at 12AM, and then suppose there are two bacteria at 12:30AM, and then there are four bacteria at 1AM, etc. (a) How many bacteria will be in the jar at 12PM that day? (b) Use trial and error to estimate at what time there will be 1 billion (1,000,000,000) bacteria in the jar. for a) is it going to be 44 bacterias ? or am i wrong? Please help with b), i have no idea how to figure it out in short term Thanks

• How many bacteria will there be at 1:30AM? (When you are stuck on a question, it is often helpful to start with an easier one.) – Jonas Meyer Jul 11 '14 at 2:56

You should notice that the number of bacteria double every 30 minutes. So, make a table:

12:00 AM - 1 bacterium

12:30 AM - 2 bacteria

01:00 AM - 4 bacteria

01:30 AM - 8 bacteria

02:00 AM - 16 bacteria

... now keep going until you hit 12 PM / 1 billion bacteria.

Here's a neat method for (b) that your teacher may discuss later, but I doubt you'd be expected to come up with this method on your own.

Using the fact that $10$ doublings multiplies by $1024$ $(1$ doubling multiplies by $2,$ $2$ doublings multiplies by $4,$ $3$ doublings multiplies by $8,\,\ldots$ keep going until you get to $10$ doublings), and $1024$ is very close to being $1000$ (a $2.4$% error), we can leap-frog by taking jumps of $10$ doublings at a time:

$$10 \;\; \text{doublings corresponds to roughly} \;\; 1000 = 10^3 \;\; \text{bacteria}$$ $$20 \;\; \text{doublings corresponds to roughly} \;\; 1000000 = 10^6 \;\; \text{bacteria}$$ $$30 \;\; \text{doublings corresponds to roughly} \;\; 1000000000 = 10^9 \;\; \text{bacteria}$$ So you'll get about $1$ billion bacteria after $30$ doublings, which corresponds to $30$ half-hour periods.

The time needed for 1 billion bacteria: Solve: $2^{(\frac{t}{30})}$ ~ $1000000000$

Counting is done at every half an hour, so in $12$ hours $24$ times counting will be done.

Number of bacteria becomes twice at each step, this tells us that it is a Geometric Progression with first term ($a$)=$1$, common ration ($r$)=2 and number of terms ($n$)=24.

Sum of all the terms for GP is given by $S=\Large\frac{a(r^n-1)}{(r-1)}$, there is only one unknown in this formula and that is $S$ to give you answer for the (a) part.

Again use this formula for (b) part, you will get $2^n=10^9+1\Longrightarrow 2^n\approx10^9$

$2^{10}=1024\approx10^3 \hspace{30 pt}\therefore2^{10}.2^{10}.2^{10}\approx10^310^310^3\hspace{30 pt}\therefore2^{30}\approx10^9$

n=number of steps/countings=$30\Rightarrow 15$hours

• i did a mistake, in part (a) use formula for the nth term of GP,=$ar^{n-1}$ – Vikram Jul 12 '14 at 8:07