Calculating Levels in a Geometric Series Phone Tree

[ThomasMagnus]

A school phone tree has 1 person responsible for contacting 3 people. If there are 1500 students in the school, how many levels will there be on the phone tree (assuming 1 person is at the top of the tree)?

My Solution:


This question forms a geometric series:
A(first term)=1
R(common ratio)=3

1+3+9+27...

Let n= # of levels to the phone tree
When will the sum of the series equal 1500?

Sn=A(1-R^n)/(1-R)

1500=1(1-(3^n))/(1-(3))

-3000=1-(3)^n

-3001=-(3^n)

3001=(3^n)

Log(3001)=nLog(3)

Log(3001)/Log(3)

~7.29

An 8th level would have to be added to the tree; however, the level would not be complete. 8 levels to the tree.

Is this the correct solution to this question? Can you identify where and if I have gone wrong?

Thanks! Appreciate the help :)

 

[eumyang]

Looks right to me. My question is, how are the levels to this tree counted? For example, IIRC, in binary trees (a data structure in computer science), the root node by itself (ie. the top of the tree) would be "Level 0." So the answer to your problem would be 7 or 8 levels, depending on whether the person at the top of the tree is counted as "Level 1" (1-2-3-4-5-6-7-8) or "Level 0" (0-1-2-3-4-5-6-7).

 

[ThomasMagnus]

I doubt there would be anything to do with computer science in this question as this is grade 12 math :)

Thanks :)