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A multiplicative inverse of 5 module 7 is?
A multiplicative inverse of 5 mod7 would be a number n ( not a unique one) such that 5n=1Let's look at the possible numbers5x1=5mode 75x2=10=3 mod 75x3=15=1 mod 7 THAT WILL DO IT3 is the multiplicative inverse of 5 mod 7.What about the others? 5x4=20, that is -1 mod 7 or 65x5=25 which is 4 mod 75x6=30 which is -5 or 2 mod 7How did we know it existed? Because 7 is a prime. For every prime number p and positive integer n, there exists a finite field with pn elements. This is an important theorem in abstract algebra. Since it is a field, it must have a multiplicative inverse. So the numbers mod 7 make up a field and hence have a multiplicative inverse.
What is remainder when 5 to the power of 30 is divided by 7?
Look at the powers of 5 mod 7: 5¹ mod 7 = 5 5² mod 7 = 5 × (5¹ mod 7) mod 7 = (5 × 5) mode 7 = 25 mod 7 = 4 5³ mod 7 = 5 × (5² mod 7) mod 7 = (5 × 4) mod 7 = 20 mod 7 = 6 5⁴ mod 7 = 5 × (5³ mod 7) mod 7 = (5 × 6) mod 7 = 30 mod 7 = 2 5⁵ mod 7 = 5 × (5⁴ mod 7) mod 7 = (5 × 2) mod 7 = 10 mod 7 = 3 5⁶ mod 7 = 5 × (5⁵ mod 7) mod 7 = (5 × 3) mod 7 = 15 mod 7 = 1 5⁷ mod 7 = 5 × (5⁶ mod 7) mod 7 = (5 × 1) mod 7 = 5 mod 7 = 5 At this point, it is obvious that the remainders will repeat the cycle {5, 4, 6, 2, 3, 1} There are 6 remainders in the cycle, so the remainder of 30 divided by 6 will tell you which remainder to use; if the remainder is 0, use the 6th element. 30 ÷ 6 = 5 r 0 →use the 6th element which is 1, so 5³⁰ ÷ 7 will have a remainder of 1. 1 ≡ 5³⁰ mod 7.
How do you get the multiplicative inverse?
Multiplicative inverses are two numbers whose product is one.Another name for multiplicative inverse is reciprocal. The reciprocal of 2/3 is 3/2. 2/3 X 3/2 = 6/6 = 1. The multiplicative inverse of 7 is 1/7. 7 X 1/7 = 7/7 = 1.
What is the unit digit in 3 power 99?
To find the units digit of 399 the question being asked is: What is (399) MOD 10? This does not necessitate evaluation of 399 before the modulus is done, as it can be done whenever it is possible during the multiplication as any multiple of 10 multiplied by 3 is still a multiple of 10. The first few powers of 3 modulus 10 are: 31 MOD 10 = 3 32 MOD 10 = (3 x 31) MOD 10 = (3 x 3) MOD 10 = 9 33 MOD 10 = (3 x 32) MOD 10 = (3 x 9) MOD 10 = 27 MOD 10 = 7 34 MOD 10 = (3 x 33) MOD 10 = (3 x 7) MOD 10 = 81 MOD 10 = 1 35 MOD 10 = (3 x 34) MOD 10 = (3 x 1) MOD 10 = 3 36 MOD 10 = (3 x 35) MOD 10 = (3 x 3) MOD 10 = 9 At this point, it can be seen that the answer is a repeating pattern of 3, 9, 7, 1, 3, 9, ... So we need the 99th element of this pattern. The pattern is a repeat of 4 digits, so we calculate 99 MOD 4 = 3. So the 3rd element of the repeating part is the answer: 7. (If the power MOD 4 had been 0, it would have been the 4th element of the pattern: 1)
What's the additive inverse of -7?
We will answers the two questions:1. What is the additive inverse of -72. What's an additive identity.The additive inverse of a number is the number you have to add to the number in order to get 0. (Or more generically speaking, to get the additive identity element of the group or field.) So the additive inverse of -7 is +7. For any real number a, the additive inverse is -a. If z is a complex number, a+bi, then the additive inverse is (-a-bi) since (a+bi)+(-a-bi)=0.The case becomes a little more interesting in fields other than the real or the complex numbers. The integers mod p, where p is a prime, form a finite field. So if we look at integers mod 7, the additive inverse of 5, for example, would be 2 since 5+2=7 which is congruent to 0 in this field.The additive identity in the field of real or complex numbers is 0."Additive identity" means the number you can add to any other number in order to get the same number back. Since -7 + 0 = -7, the additive identity of -7 is 0.In the case of a+bi where i^2=-1, the additive identity is still 0. If it helps you to think of it as 0+0i, that is fine. In the finite field of integers mod p, where p is a prime, we have p as the additive identity. For example, 2 mod 7 is just 2, and if we add 7 it is 9 but that is still 2 mod 7.All of these ideas can be extended to fields of invertible matrices and many other exciting algebraic structures!
A multiplicative inverse of 5 module 7 is?
A multiplicative inverse of 5 mod7 would be a number n ( not a unique one) such that 5n=1Let's look at the possible numbers5x1=5mode 75x2=10=3 mod 75x3=15=1 mod 7 THAT WILL DO IT3 is the multiplicative inverse of 5 mod 7.What about the others? 5x4=20, that is -1 mod 7 or 65x5=25 which is 4 mod 75x6=30 which is -5 or 2 mod 7How did we know it existed? Because 7 is a prime. For every prime number p and positive integer n, there exists a finite field with pn elements. This is an important theorem in abstract algebra. Since it is a field, it must have a multiplicative inverse. So the numbers mod 7 make up a field and hence have a multiplicative inverse.
Why does -24 mod 7 equal 4 but 24 mod 7 equal 3?
The modulus of a number relative to a base is the positive remainder when itis divided by that base. So -24 mod 7 = (-28 + 4) mod 7 = -28 mod 7 + 4 mod 7 = 4 mod 7 = 4while 24 mod 7 = (21 + 3) mod 7 = 21 mod 7 + 3 mod 7 = 3 mod 7 = 3.
What is the multiplicative inverse of 2 and one third?
2 1/3 = 7/3 inverse of 7/3 is 3/7
How do you the multiplicative inverse of -3 and 3 seventh?
The multiplicative inverse of -24/7 is -7/24
What is the inverse for the equation 7x plus 3?
To find the inverse of a function, simply switch the variables x and y. So for the function y=7x+3, the inverse would be x=7y+3, or y=(x-3)/7.
What is the multiplicative inverse of 3 and a half?
3 and a half = 7/2 so its multiplicative inverse is 2/7.
What is remainder when 5 to the power of 30 is divided by 7?
Look at the powers of 5 mod 7: 5¹ mod 7 = 5 5² mod 7 = 5 × (5¹ mod 7) mod 7 = (5 × 5) mode 7 = 25 mod 7 = 4 5³ mod 7 = 5 × (5² mod 7) mod 7 = (5 × 4) mod 7 = 20 mod 7 = 6 5⁴ mod 7 = 5 × (5³ mod 7) mod 7 = (5 × 6) mod 7 = 30 mod 7 = 2 5⁵ mod 7 = 5 × (5⁴ mod 7) mod 7 = (5 × 2) mod 7 = 10 mod 7 = 3 5⁶ mod 7 = 5 × (5⁵ mod 7) mod 7 = (5 × 3) mod 7 = 15 mod 7 = 1 5⁷ mod 7 = 5 × (5⁶ mod 7) mod 7 = (5 × 1) mod 7 = 5 mod 7 = 5 At this point, it is obvious that the remainders will repeat the cycle {5, 4, 6, 2, 3, 1} There are 6 remainders in the cycle, so the remainder of 30 divided by 6 will tell you which remainder to use; if the remainder is 0, use the 6th element. 30 ÷ 6 = 5 r 0 →use the 6th element which is 1, so 5³⁰ ÷ 7 will have a remainder of 1. 1 ≡ 5³⁰ mod 7.
Additive inverse property?
Additive inverse just means that when you add them, it equals 0. So, 1 and -1; 12 and -12; -74 and 74; 0 and -0. It can also be applied to things other than integers: x+12 and -x-12; 2x-3 and -2x+3 (which is the same as 3-2x) Or, if you know anything about modular arithmetic, then: 7+x≡0 (mod 9) and 2; 6-x≡4 (mod 7) and -2
What is the multiplicative inverse of 3 1 2?
3 1/2 = 7/2 in improper fractionmultiplicative inverse is the reciprocal of 7/2 = 2/7
How do you turn 5 and 3-7ths into a multiplicative and a additive inverse?
38/7 x 7/38 = 1 = multiplicative inverse 38/7 + -38/7 = 0 = additive inverse
How do you find the multiplicative inverse of a fraction?
Swap the numerator and denominator. For example, the multiplicative inverse of 5/7 is 7/5
How do you find the multiplicative inverse of a whole number?
Divide 1 by the number. The multiplicative inverse of 7 is 1/7, for example.
