A community organizes a phone tree in order to alert each family of emergencies. In the first stage, one person calls five families. In the second stage, each of the five families calls another five families, and so on. How many stages need to be reached before 19,530 families or more have been called?

Question

A community organizes a phone tree in order to alert each family of emergencies.  In the first stage, one person calls five families.  In the second stage, each of the five families calls another five families, and so on.  How many stages need to be reached before 19,530 families or more have been called?

anonymous · Answer

More formulas to be used. Is it recursive?

anonymous · Answer

$$a_{n} = (5)(19530)^{n - 1}$$

ddcamp · Answer

Not quite recursive, but you would have the sum:
$$\sum_{n=0}^{A} 5^n = 1 + 5 + 25 + 125 + 625 +... +5^A$$

anonymous · Answer

(5n + 1) = 1 + 5 + 25 + 125 + 625? 781 stages?

anonymous · Answer

5n - 780 = 0
n = 156?

ddcamp · Answer

At the 0 stage, 1 person has been "contacted". At the 1st stage, 6 people (the 5 new and the original) have been contacted. At the second stage, the first 6 and 25 new people have been contacted.

on the nth stage, you have the the people from before plus 5^n new people.

Which power of 5 is closest to 19530?

phi · Answer

use the sum of a geometric series https://en.wikipedia.org/wiki/Geometric_series#Formula

phi · Answer

in your case
$$\sum_{k=0}^{N}ar^k = a \frac{ 1-r^{N+1 }}{ 1-r }$$
a=1 and r= 5
and you want to solve for N

phi · Answer

in other words, solve for N in
$$ \frac{1 -5^{N+1}}{-4}= 19530 $$

anonymous · Answer

-3125?

phi · Answer

the first step is multiply both sides by -4
$$ -4 \cdot \frac{1 -5^{N+1}}{-4}= 19530 \cdot -4 \
1 -5^{N+1} = - 78120 
$$
add -1 to both sides
$$ -1 + 1 - -5^{N+1} = - 78120 -1 \-5^{N+1} = - 78121$$
multiply both sides by -1
$$ 5^{N+1} =  78121$$

phi · Answer

to find N+1, take the log of both sides
$$ \log(5^{N+1}) = \log(78121) $$

phi · Answer

use the property 
$$ \log(a^b) = b \log(a) $$

phi · Answer

$$ \log(5^{N+1}) = \log(78121) \ (N+1) \log(5) = \log(78121) $$
divide both sides by log 5
$$ N+1 = \frac{\log(78121)}{\log(5) } $$
add -1 to both sides
$$ N=  \frac{\log(78121)}{\log(5) } -1 $$

phi · Answer

you will need a calculator to find N

anonymous · Answer

5.9999
n = 6?
$$\frac{ 7815 }{ 2 }$$

phi · Answer

yes, n=6

anonymous · Answer

$$3907\frac{ 1 }{ 2 }$$

phi · Answer

what is 3907 ½ ?

anonymous · Answer

Mixed fraction of 7815. 6 stages need to be reached before 19,530?

anonymous · Answer

How can I double check that?

anonymous · Answer

$$\frac{1 -5^{6+1}}{-4}$$

phi · Answer

In the first stage, one person calls five families.
that means
1+5  know after 1 stage
In the second stage, each of the five families calls another five families,
that means 1+ 5 + 25 know after 2 stages

6 stages will have 7 numbers:

1+5+25+125+625+3125+15625

that we add up

phi · Answer

If we did not use the equation, we could solve the problem by keep adding stages and adding up as we go  until we pass 19530

anonymous · Answer

It's like a shortcut instead of adding all those numbers up?

phi · Answer

yes.  For 7 numbers we don't really need it.. but it is a good formula to know.

anonymous · Answer

Thank you @DDCamp as well.