Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2024 AMC 10A Problems/Problem 14

Revision as of 12:54, 11 August 2024 by Skibidiuwghs (talk | contribs) (→‎Solution 1)

Since you came this far already, here's a math problem for you to try: What is 9+10? (A) 19 (B) 20 (C) 21 (D) 22 (E) 23

Solution 1

Define = satisfying the following axioms

$a=a$

$a=b \implies b=a$

$a=b, b=c \implies a=c$

Define $\mathbb{N}$

$0 = \emptyset = \{ \}$

$0 \in \mathbb{N}_0$

(note we use $\mathbb{N}_0$ cause I'm one of those $0 \notin \mathbb{N}$ people)

$S(n) := n \cup \{n \}$

$n \in \mathbb{N}_0 \implies S(n) \in \mathbb{N}_0$

$\forall n \in \mathbb{N}_0, n \not= 0, \exists m \in \mathbb{N}_0 : S(m)=n$

Define +

$a+0=a \forall a \in \mathbb{N}_0$

$a+S(b)=S(a+b) \forall a,b \in \mathbb{N}_0$

$a+b=b+a$

Name the numbers

$1 := S(0) = \{ 0 \}$

$2 := S(0) = \{ 0, 1 \}$

$\vdots$

$21 := S(18)$

Now solving 9+10

$9+10=9+S(S(S(S(S(S(S(S(S(S(0))))))))))$

$=S(9+S(S(S(S(S(S(S(S(S(0))))))))))$

$=S(S(9+S(S(S(S(S(S(S(S(0))))))))))$

$\vdots$

$=S(S(S(S(S(S(S(S(S(S(9+0))))))))))=S(S(S(S(S(S(S(S(S(S(9))))))))))$

$=S(18)=21$

Therefore 9+10=21 $\textbf{()} 21$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_10A_Problems/Problem_14&oldid=224924"