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

Revision as of 12:55, 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 +

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

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

$a+b=b+a$

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 $\textbf{(C)} 21$

Revision as of 12:54, 11 August 2024 (view source) Skibidiuwghs (talk \| contribs) (→‎Solution 1) ← Older edit		Revision as of 12:55, 11 August 2024 (view source) Skibidiuwghs (talk \| contribs) (→‎Solution 1) Newer edit →
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	<math>=S(18)=21</math>		<math>=S(18)=21</math>

−	Therefore 9+10=21 <math>\textbf{()} 21</math>	+	Therefore 9+10=21 <math>\textbf{(C)} 21</math>