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1) Suppose n is a positive number between 1 and 11 inclusive, and \(n^2 \equiv 0 \pmod{12}.\)What is n?

 

2) Grogg skip counts by 11s starting at 12. He obtains the sequence: 12, 23, 34, ..., 1233. What is the remainder when the sum of all the terms in Grogg's sequence is divided by 11?

 Oct 20, 2019
 #1
avatar+437 
+5

1) http://mathworld.wolfram.com/ModularArithmetic.html 

2)https://web2.0calc.com/questions/modular-arithmetic_3

 

use these links for reference.

 Oct 21, 2019
 #2
avatar+2547 
+1

Only use the hint below if you ARE SUPER STUCK and you can't solve it at all.

 

Here is a hint for 1.

 

If the n is divisible by 12, that means it is also divisible by 3 and 4.

 

That also means it is divisible by 3 and 2

 

What number between 1 and 11 is divisible by 3 and 2?

CalculatorUser  Oct 21, 2019
 #3
avatar+23903 
+1

1)
Suppose n is a positive number between 1 and 11 inclusive, and \(n^2 \equiv 0 \pmod{12}\).
What is \(n\)?

 

\(\begin{array}{|rcll|} \hline n^2 &\equiv & 0 \pmod{12} \quad | \quad 0 \equiv 12 \equiv 24 \equiv 36 \pmod{12} \quad 36\text{ is a perfect square}\\ n^2 &\equiv & 36 \pmod{12} \quad | \quad sqrt() \\ n &\equiv & \pm \sqrt{36} \pmod{12} \\ \hline && \begin{array}{|rclcrcl|} \hline n &\equiv & 6 \pmod{12} &or & n &\equiv & -6 \pmod{12} \\ & & && n &\equiv & -6+12 \pmod{12} \\ & & && n &\equiv & 6 \pmod{12} \\ \hline \end{array} \\ \hline \end{array}\)

 

\(n\) is 6

 

laugh

 Oct 21, 2019
edited by heureka  Oct 21, 2019
 #4
avatar+23903 
+1

2)
Grogg skip counts by 11s starting at 12. He obtains the sequence: \(12, 23, 34, \ldots, 1233\).
What is the remainder when the sum of all the terms in Grogg's sequence is divided by 11?

 

\(\begin{array}{|rcll|} \hline &&\mathbf{ 12 + \underbrace{(12+11)}_{23} + \underbrace{(12+2\cdot 11)}_{34}+ \ldots + \underbrace{(12+(n-1)\cdot 11)}_{1233} \pmod{11} }\\ &\equiv& 12 +(12+0) + (12+0)+ \ldots + (12+0) \pmod{11} \\ &\equiv& 12n \pmod{11} \quad | \quad 12\pmod{11} = 1 \\ &\equiv& 1n \pmod{11} \\ &\equiv& n \pmod{11} \\ && \boxed{ 12+(n-1)\cdot 11 = 1233\\ 12+11n-11 =1233\\ 1+11n = 1233\\ 11n = 1232\\ n=\dfrac{1232}{11}\\ n = 112 } \\ &\equiv& 112 \pmod{11} \\ &\equiv& 110 + 2 \pmod{11} \quad | \quad 110 \pmod{11} = 0 \\ &\equiv& 0 + 2 \pmod{11} \\ &\equiv& \mathbf{ 2 \pmod{11} } \\ \hline \end{array} \)

 

The remainder when the sum of all the terms in Grogg's sequence is divided by 11 is 2

 

laugh

 Oct 21, 2019
 #5
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0

Hello Guest! I suppose you are one of the people who attend Middle School Competition Math! I would kindly advise not to post the questions on here, unless you want to be scolded by the director! Thank you, and have a good day.

 Oct 21, 2019
 #6
avatar
0

I am very sorry, I actually did not mean to post this question. I will never do that again. Sorry... sad

Guest Oct 21, 2019
 #8
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0

Now who might you be tho? I ask thee this question.

Guest Oct 22, 2019
 #7
avatar
0

I think the first one is 6, but I'm not sure, so don't take my word for it.

 Oct 21, 2019

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