3x = 1 (mod 5) is determined when their units is 1 or 6.
By using trial and error, we got :
2,7,12,17,...
We must find the pattern.
Use the formula :
An = b(n-1)+a
a is the 1st term, so a = 2.
b is the difference, so b = 5
An = 5(n-1)+2 = 5n - 5 + 2 = 5n - 3
So, when n is a natural number, x=5n-3 is a solution to 3x=1(mod5)
2007-10-16 00:44:52
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answer #1
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answered by wangsacl 4
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I don't know how to solve congruences the easy way,
so I have to go back to basics.
3x ≡ 1 (mod 5)
This means that (3x - 1) is exactly divisible by 5.
So we can let an integer, J = (3x - 1) / 5
Therefore, x = (5J + 1) / 3
Dividing as far as we can gives : x = J + (2J + 1) / 3
These are all integers, so let K = (2J + 1) / 3
Therefore, J = (3K - 1) / 2
Again, dividing as far as we can gives : J = K + (K - 1) / 2
All integers again, so let M = (K - 1) /2
Therefore, K = 2M + 1. Now we are rid of fractions.
Substituting back, with K = 2M + 1 gives :
J = 3M + 1
and back further gives : x = 5M + 2
Now let M take on the values 0, 1, 2, 3, 4, 5, ...
Thus, x = 2, 7, 12, 17, 22, 27, ...
So the solution is : x = 5M + 2
for M = 0, 1, 2, 3, ...
2007-10-16 02:13:38
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answer #2
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answered by falzoon 7
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3x - 5 = 1
3x = 6
x = 2
thats the first value of x,
just add 5 the next values, thats is
2nd value = 2 + 5 = 7
check : 3*7 = 21 = 1 (mod 5)
3rd value = 7 + 5 = 12
check : 3 * 12 = 36 = 1 (mod 5)
4th value = 12 + 5 = 17
check: 3 * 17 = 51 = 1 (mod 5)
5th value: 17 + 5 = 22
check: 3 * 22 = 66 = 1 ( mod 5)
and so on..
2007-10-16 01:11:53
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answer #3
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answered by jen ol 3
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Solving Modulus Equations
2016-11-13 00:33:37
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answer #4
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answered by ? 4
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I think you mean 3 times x is congruent to 1 modulo 5. So, we have to find the positive integers x such that 5 divides 3x -1.This happens if, and only if, 3x -1 ends in 0 or 5, that is, if, and only if, 3x ends in 1 or in 6.
3x ends in 1 if, and only if, x ends in 7, that is, x s a number like 7, 17, 27...So, the numbers of the form x = 7 + 10(n-1)= 10n -3.
3x ends in 6 if, and only if, x ends in 2, that is, if and only if x is even.
So, the solution to your congruence is the set {x | x = 10n -3 or x = 2n, n positive integer}
2007-10-16 00:59:51
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answer #5
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answered by Steiner 7
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Did you mean 3 times x or 3 to the power of?
2007-10-16 00:23:03
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answer #6
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answered by Ali 3
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x=0 coz any number with the power of zero is one!
or if u mean 3x well, srry i dont know!
2007-10-16 00:27:57
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answer #7
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answered by Anonymous
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