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

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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
 +
 +
==Solution 1==
 +
Define = satisfying the following axioms
 +
 +
<math>a=a</math>
 +
 +
<math>a=b \implies b=a</math>
 +
 +
<math>a=b, b=c \implies a=c</math>
 +
 +
Define <math>\mathbb{N}</math>
 +
 +
<math>0 = \emptyset = \{ \}</math>
 +
 +
<math>0 \in \mathbb{N}_0</math>
 +
 +
(note we use <math>\mathbb{N}_0</math> cause I'm one of those <math>0 \notin \mathbb{N}</math> people)
 +
 +
<math>S(n) := n \cup \{n \}</math>
 +
 +
<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>
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 +
Define +
 +
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<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>
 +
 +
Name the numbers
 +
 +
<math>1 := S(0) = \{ 0 \} </math>
 +
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<math>2 := S(0) = \{ 0, 1 \} </math>
 +
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<math>\vdots</math>
 +
 +
<math>21 := S(18)</math>
 +
 +
Now solving 9+10
 +
 +
<math>9+10=9+S(S(S(S(S(S(S(S(S(S(0))))))))))</math>
 +
 +
<math>=S(9+S(S(S(S(S(S(S(S(S(0))))))))))</math>
 +
 +
<math>=S(S(9+S(S(S(S(S(S(S(S(0))))))))))</math>
 +
 +
<math>\vdots</math>
 +
 +
<math>=S(S(S(S(S(S(S(S(S(S(9+0))))))))))=S(S(S(S(S(S(S(S(S(S(9))))))))))</math>
 +
 +
<math>=S(18)=21</math>
 +
 +
Therefore 9+10=21 <math>\textbf{(C)} 21</math>

Latest revision as of 13:55, 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

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 $\textbf{(C)} 21$

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