Solving a TI-83/84 Calculator Exam Question: 771 vs. 1,059

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The discussion centers on solving a TI-83/84 calculator exam question regarding family attendance at a meeting since 1961. The problem states that a family started with 71 members and grows annually by nearly 6% due to various factors, with 39 members missing the 2008 meeting. Two methods were used to calculate total attendance: one through calculator programming, yielding 771 members, and the other using a compound interest formula, resulting in 1,059 members. The second method is considered more logical, as the programming approach lacks clarity without sharing specific lines of code. Ultimately, the discrepancy between the two answers highlights the importance of method selection in mathematical problem-solving.
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This is a question from a TI-83/84 calculator exam that I took on February 14th. (haha what a shocker a math nerd without a girlfriend.) Anyway, this is one of the few that I missed, and when I revisited it I noticed that I had actually misread the question at hand. Anyway, the problem is that the two ways that I have tried to solve the problem have given me vastly different answers.

1.
A family has met on September 1st to visit since 1961. At that time, there were 71 family members, and each year the family grew as a net result of marriages, divorces, adoptions, births, and deaths by an amount equal to the largest integer less that 6% of the total from the previous year. There were 39 family members who missed the 2008 meeting. How many attended the 2008 meeting?

2.
The first included programing the calculator which gave me an answer of 771 total family members. Umm, I guess I could post the programming lines? Ask if you do!

The second was a different take on the problem, in which I instead set it up like one would if trying to find compound interest. That gave me: 1,059 total family members.

1,059 = 71(1 + .06/1)^(1*47)

3.
I've already stated my answers above, but here they are again:
Calculator Program: 771 total family members.
Compound Interest: 1,059 total family members.
 
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I went about it the second way, too.

missbooty87

(oh, and math nerds are not only guys. they are girls, too. And I was without a boyfriend on feb 14 ha ha ha i think its a nerd thing)
 
haha Sorry to leave out the fairer side of the math nerds or really nerds in general! haha so yeah, The second way makes more sense I think.
 
Especially since you haven't shown us what you did in programming your calculator.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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