What are the 3 solutions? I'm stuck! 6x=15(mod 21)
a=6,m=21,b=15 d=gcd(6,21)=3 solns.
6x=15(mod 21) 2x=5(mod 7)
21=6(3)+3
6+3(2)+0
0=6-6
6-(21-6(3))=3
6-21+18=3
6( )-21()=3
(21-6(3))(15)-6=3 ???????
To find the solutions to the equation 6x ≡ 15 (mod 21), we can first simplify it by dividing both sides by the greatest common divisor (gcd) of 6 and 21, which is 3.
Step 1: Simplifying the equation
Divide both sides of the equation by 3:
(6x)/3 ≡ 15/3 (mod 21)
2x ≡ 5 (mod 21)
Step 2: Simplifying the modulus
Since 21 = 3 * 7, we can rewrite the equation as:
2x ≡ 5 (mod 3*7)
Step 3: Solving the simplified equation
We can now solve the equation 2x ≡ 5 (mod 7):
To find the inverse of 2 modulo 7, we can try values of x such that 2x/7 is an integer. By computing, we can see that 2 * 4 ≡ 8 ≡ 1 (mod 7), so the inverse of 2 modulo 7 is 4.
Multiply both sides of the equation by 4:
4 * 2x ≡ 4 * 5 (mod 7)
8x ≡ 20 (mod 7)
Simplify the equation:
x ≡ 6 (mod 7)
Therefore, the equation 6x ≡ 15 (mod 21) has the following solutions:
x ≡ 6 (mod 7)
Please note that x can also take on values such as x ≡ 6 + 7k, where k is an integer, since we are working in modular arithmetic.
To find the solutions to the equation 6x ≡ 15 (mod 21), you can use the method of modular arithmetic and the concept of the multiplicative inverse.
1. First, find the greatest common divisor (gcd) of 6 and 21 using the Euclidean algorithm.
gcd(6, 21) = 3.
2. Since the gcd is not equal to 1, there are three solutions to the equation.
3. To simplify the equation, divide both sides by the gcd.
(6/3)x ≡ (15/3) (mod 21).
2x ≡ 5 (mod 21).
4. Now, you need to find the multiplicative inverse of 2 modulo 21. This is a number that, when multiplied by 2, gives a remainder of 1 when divided by 21.
5. One way to find the multiplicative inverse is by using the extended Euclidean algorithm. However, this method can be a bit complex. Alternatively, you can use trial and error.
6. Try multiplying 2 by different numbers and finding the remainder when divided by 21. Keep trying until you find a number that gives a remainder of 1.
7. Starting from 2, try 2 * 1 = 2 (remainder 2), 2 * 2 = 4 (remainder 4), 2 * 3 = 6 (remainder 6), 2 * 4 = 8 (remainder 8), and so on.
8. Keep trying until you find a remainder of 1. In this case, you will find that 2 * 11 = 22 (remainder 1) when divided by 21. So, the multiplicative inverse of 2 modulo 21 is 11.
9. Multiply both sides of the equation by the multiplicative inverse.
11 * 2x ≡ 11 * 5 (mod 21).
22x ≡ 55 (mod 21).
10. Simplify the equation.
x ≡ 55 (mod 21).
11. Now, since the modulus is 21, the solutions for x should be between 0 and 20.
12. Calculate the values of x that satisfy the equation. Start from x = 0 and keep adding 21 until you find a solution.
x = 0: 55 ≡ 55 (mod 21) - This is a solution.
x = 21: 76 ≡ 11 (mod 21) - This is not a solution.
x = 42: 97 ≡ 34 (mod 21) - This is not a solution.
...
13. Continue this process until you find three distinct solutions that satisfy the equation.
So, the three solutions to the equation 6x ≡ 15 (mod 21) are x ≡ 0 (mod 21), x ≡ 11 (mod 21), and x ≡ 16 (mod 21).
6x = 21y + 15 for some y.
or,
2x = 7y + 5
since 2x is even, y must be odd.
y x
1 6
3 13
5 20
7 27
. . .
any number x of the form 2x = 7(2k+1)+5 is a solution, since
2x = 14k+7+5 = 14k+12
6x = 42k + 36
and 36 = 15 mod 21