Difference between revisions of "2024 AMC 10A Problems/Problem 14"

Revision as of 12:40, 11 August 2024

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

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 +

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

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

$a+b=b+a$

@@ Line 5: / Line 5: @@
 ==Solution 1==
 Define = satisfying the following axioms
 <math>a=a</math>
@@ Line 24: / Line 25: @@
 <math>\forall n \in \mathbb{N}_0, n \not= 0, \exists m \in \mathbb{N}_0 : S(m)=n</math>
+Define +
+<math>a+0=a \forall a \in \mathbb{N}_0</math>
+<math>a+S(b)=S(a+b) \forall a,b \in \mathbb{N}_0</math>
+<math>a+b=b+a</math>