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avatar+1148 

Find the largest negative integer x which satisfies the congruence \(34x+6\equiv 2\pmod {20}\).

tertre  Mar 23, 2017
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6+0 Answers

 #1
avatar+6765 
0

If 34x + 6 congruent 2 mod 20, 34x + 4 is divisible by 20.

 

We use trial and error(again) and test for each negative integer from -1.

 

34(-1) + 4 = -30 which is not divisible by 20.

34(-2) + 4 = -60<-- OMG answer already here. It is divisible by 20.

Therefore the largest negative integer which satisfies the congruence is -2 :)

 

~The smartest cookie in the world.

MaxWong  Mar 23, 2017
 #6
avatar+310 
0

Nope.

 

34*(-2)+4=30*(-2)+4*(-2)+4=30*(-2)-4=-64

 

-64 is not divisible by 20.

 

:/

 

heres my answer:

 

34x+4 is divisble by 20

meaning

17x+2 is divisble by 10

meaning

7x+2 is divisble by 10.

 

All you have to do is to calculate until you get to the right number

Ehrlich  Mar 24, 2017
 #7
avatar+6765 
0

Well..... I am not a very careful person and I made a mistake.....

MaxWong  Mar 26, 2017
 #3
avatar+1148 
0

Isn't it -3?

tertre  Mar 23, 2017
 #4
avatar+6765 
0

-2 is larger than -3.

MaxWong  Mar 23, 2017
 #5
avatar+18377 
+2

Find the largest negative integer x which satisfies the congruence

\( 34x+6\equiv 2\pmod {20}\).

 

\(x = -6+10n \quad | \quad n \in \mathbb{Z}\)

 

The largest negative integer x = - 6

 

Check:

\(\begin{array}{|rcll|} \hline && 34\cdot(-6)+6 \pmod {20} \\ &\equiv & -204 + 6 \pmod {20} \\ &\equiv & -198 \pmod {20} \\ &\equiv & -18 \pmod {20} \\ &\equiv & -18+20 \pmod {20} \\ &\equiv & 2 \pmod {20} \\ \hline \end{array}\)

 

laugh

heureka  Mar 24, 2017

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