## Problem 1

1. $7^3$
2. $5 + 4^2$
3. $1213 - 9^3$

$7^3$ $5 + 4^2$ $1213 - 9^3$
$7 \times 7 \times 7$ $5 + 4 \times 4$ $1213 - 9 \times 9 \times 9$
$49 \times 7$ Multiply the 4s first Multiply the 9s first
343 $5 + 16$ $1213 - 729$
21 484

## Problem 2

2.a $10^4$
2.b $10^7$
2.c $10^10$

2.a $10^4 = 10 \times 10 \times 10 \times 10 = 10,000$
2.b $10^7 = 10 \times 10 \times 10 \times 10 \times 10 \times 10 \times 10 = 10,000,000$
2.c $10^10 = 10,000,000,000$

## Problem 3

Everybody is born to $2^1$ biological parents. Our parents each had $2^1 + 2^1$ biological parents. We can say that our grandparents are $2^2$ mathematically as the number of our ancestors doubles with each generation we go back.
So:
3.a How many times would 2 be multiplied to determine the number of great grandparents?
3.b How many times would 2 be multiplied to determine the number of great-great grandparents?
3.c How many people would be our 28 ancestors?

3.a If our grandparents are the 22 generation, then our great-grandparents are one more back so are 23.

3.b This means our great-great grandparents are one more generation back so would be our 24 ancestors.

3.c Our 28 ancestors would be $2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 256$ people.

## Problem 4

Problem from previous page

You should have 32, or $2^5,$ scraps of paper left over. The general solution is:
$P = 2^r$, where P is the number of paper scraps and r is the number of times the paper has been torn.
An interesting anecdote related to this problem: it was once commonly believed that a piece of paper could only ever be folded 8 times; thus, when unraveled, the paper could contain at most 256 ($2^8$) sections. However, this was later proven false when Britney Gallivan folded a paper 12 times, and derived a function for the actual number of folds that could occur.