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

(added a joke (other pages for 2024amc10a have jokes on them too))
 
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What is 9+10?
 
What is 9+10?
 
(A) 19 (B) 20 (C) 21 (D) 22 (E) 23
 
(A) 19 (B) 20 (C) 21 (D) 22 (E) 23
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 +
==Solution 1==
 +
Define = satisfying the following axioms
 +
<math>a=a</math>
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<math>a=b \implies b=a</math>
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<math>a=b, b=c \implies a=c</math>
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Define <math>\mathbb{N}</math>
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<math>0 = \emptyset = \{ \}</math>
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<math>0 \in \mathbb{N}_0</math>
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(note we use <math>\mathbb{N}_0</math> cause I'm one of those <math>0 \notin \mathbb{N}</math> people)
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<math>S(n) := n \cup \{n \}</math>
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<math>n \in \mathbb{N}_0 \implies S(n) \in \mathbb{N}_0</math>
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<math>\forall n \in \mathbb{N}_0, n \not= 0, \exists m \in \mathbb{N}_0 : S(m)=n</math>

Revision as of 12:36, 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

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

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