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# mathquestionxtw

+2
290
7
+2720

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

tertre  Mar 23, 2017
#1
+6979
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
+310
0

Nope.

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

-64 is not divisible by 20.

:/

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
+6979
0

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

MaxWong  Mar 26, 2017
#3
+2720
0

Isn't it -3?

tertre  Mar 23, 2017
#4
+6979
0

-2 is larger than -3.

MaxWong  Mar 23, 2017
#5
+19496
+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:

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

heureka  Mar 24, 2017