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Middle School Math Grades 5-8, Ages 10-13, MATHCOUNTS, AMC 8

Grades 5-8, Ages 10-13, MATHCOUNTS, AMC 8

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Middle School Math Grades 5-8, Ages 10-13, MATHCOUNTS, AMC 8

Grades 5-8, Ages 10-13, MATHCOUNTS, AMC 8

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574 posts

maxamc

#1 h Apr 9, 2025, 3:05 PM • 1 Y

Y by PikaPika999

If 1+1=2, 2+2=4, 3+3=6, 4+4=?

Also prove $3 \nmid 7$ .

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#2 h Apr 9, 2025, 3:14 PM

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EDIT: oh it wasn't a trick question

This post has been edited 2 times. Last edited by PikaPika999, Apr 9, 2025, 3:29 PM

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#3 h Apr 9, 2025, 3:16 PM • 2 Y

Y by PikaPika999, Exponent11

We can obviously see that the variables on the left are arbitrary, so the next part of the sequence $2, 4, 6$ is $\boxed{48}$ through the polynomial $x^3 - 6x^2 + 13x - 6$ .

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evt917

#4 h Apr 9, 2025, 3:29 PM • 1 Y

Y by PikaPika999

SpeedCuber7 wrote:

We can obviously see that the variables on the left are arbitrary, so the next part of the sequence $2, 4, 6$ is $\boxed{48}$ through the polynomial $x^3 - 6x^2 + 13x - 6$ .

we can obviously see that this approach is ~~insane~~

and probably took 5 hours to come up with

jk

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778 posts

#5 h Apr 9, 2025, 3:29 PM • 1 Y

Y by PikaPika999

For 7/3 you just need to use Euclidean Algorithm and prove that gcd(3,7)=1 ez

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#6 h Apr 9, 2025, 3:53 PM • 1 Y

Y by PikaPika999

xHypotenuse wrote:

For 7/3 you just need to use Euclidean Algorithm and prove that gcd(3,7)=1 ez

This proof is incomplete. For instance, the following is false: $1 \nmid 7$ because $\text{gcd} (1,7)=1$ . Thus, we must find another way to complete the proof.

Firstly, we may apply the following lemma: if $a \mid b$ , then there exists an integer $k$ such that $ak=b$ . So, applying the following lemma we must determine whether or nor $\tfrac{7}{3} \in \mathbb{Z}$ . Firstly, we must determine whether or not $\tfrac{7}{3}$ is rational, for which we refer to the continued fraction representation of $\tfrac{7}{3}$ .
We note the following: a number is rational if and only if it has a finite canonical continued fraction representation.
Thus, we have that since $\tfrac{7}{3}=2+\tfrac{1}{3}$ as our continued fraction representation, $\frac{7}{3}$ is indeed rational. However, we also note that $\tfrac{1}{3}$ is rational. Now, we note the following:
$a,b$ are not divisible by $3$ and are not equivalent $\pmod{3}$ if and only if $3 \mid a+b$ . Thus, if we allow $a=7$ and $b=1$ (essential as these are the results from the application of the first lemma), we have $3 \mid 8$ . However, this is clearly false, since $3 \nmid 2^3$ . Thus, $7 \equiv 1 \pmod{3}$ and consequently, since $3 \nmid 1$ , we must have $3 \nmid 7$ . $\square$

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#7 h Apr 9, 2025, 4:00 PM • 1 Y

Y by PikaPika999

SirAppel wrote:

xHypotenuse wrote:

For 7/3 you just need to use Euclidean Algorithm and prove that gcd(3,7)=1 ez

This proof is incomplete. For instance, the following is false: $1 \nmid 7$ because $\text{gcd} (1,7)=1$ . Thus, we must find another way to complete the proof.

Firstly, we may apply the following lemma: if $a \mid b$ , then there exists an integer $k$ such that $ak=b$ . So, applying the following lemma we must determine whether or nor $\tfrac{7}{3} \in \mathbb{Z}$ . Firstly, we must determine whether or not $\tfrac{7}{3}$ is rational, for which we refer to the continued fraction representation of $\tfrac{7}{3}$ .
We note the following: a number is rational if and only if it has a finite canonical continued fraction representation.
Thus, we have that since $\tfrac{7}{3}=2+\tfrac{1}{3}$ as our continued fraction representation, $\frac{7}{3}$ is indeed rational. However, we also note that $\tfrac{1}{3}$ is rational. Now, we note the following:
$a,b$ are not divisible by $3$ and are not equivalent $\pmod{3}$ if and only if $3 \mid a+b$ . Thus, if we allow $a=7$ and $b=1$ (essential as these are the results from the application of the first lemma), we have $3 \mid 8$ . However, this is clearly false, since $3 \nmid 2^3$ . Thus, $7 \equiv 1 \pmod{3}$ and consequently, since $3 \nmid 1$ , we must have $3 \nmid 7$ . $\square$

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get out

well the euclidean algorithm approach works when the denominator isn't 1 though

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Craftybutterfly

478 posts

Craftybutterfly

#8 h Apr 9, 2025, 4:43 PM

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got $1$ $~~~~~~~~~~~~~~~$

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#9 h Apr 9, 2025, 4:54 PM

Y by

evt917 wrote:

SpeedCuber7 wrote:

We can obviously see that the variables on the left are arbitrary, so the next part of the sequence $2, 4, 6$ is $\boxed{48}$ through the polynomial $x^3 - 6x^2 + 13x - 6$ .

we can obviously see that this approach is ~~insane~~

and probably took 5 hours to come up with

jk

what do you mean it's just the ~~absolutely un~~intended way to solve the problem

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Existing_Human1

212 posts

Existing_Human1

#10 h Apr 9, 2025, 7:43 PM

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For $3 \nmid 7$

Assume that $3 \mid 7$ , then $7 = 3n$ where $n \in \mathbf{N}$ , then $1434 \mid 7$ , which is clearly not true, since 7 is a lucky number, and you can never lose 7 times, so QED

This post has been edited 1 time. Last edited by Existing_Human1, Apr 9, 2025, 7:43 PM

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#11 h Apr 11, 2025, 4:22 AM

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For 4+4, the answer is clearly 9, only an idiot would answer 8

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