First of all, the modular inverse of n modulo k can only exist if GCD(n, k) = 1.
We have
130 = 2 • 5 • 13
231 = 3 • 7 • 11
so n must be free of 2, 3, 5, 7, 11, and 13, which are the first six primes. It follows that n = 17 must the least integer that satisfies the conditions.
To verify the claim, we try to solve the system of congruences
Use the Euclidean algorithm to express 1 as a linear combination of 130 and 17:
130 = 7 • 17 + 11
17 = 1 • 11 + 6
11 = 1 • 6 + 5
6 = 1 • 5 + 1
⇒ 1 = 23 • 17 - 3 • 130
Then
23 • 17 - 3 • 130 ≡ 23 • 17 ≡ 1 (mod 130)
so that x = 23.
Repeat for 231 and 17:
231 = 13 • 17 + 10
17 = 1 • 10 + 7
10 = 1 • 7 + 3
7 = 2 • 3 + 1
⇒ 1 = 68 • 17 - 5 • 231
Then
68 • 17 - 5 • 231 ≡ = 68 • 17 ≡ 1 (mod 231)
so that y = 68.
THE PARABOLA OPENS DOWN (WHICH IS MAXIMUM ) AND X=1/2
Answer:
i cant see the question
Step-by-step explanation:
Answer:
The equation that models this situation is:
x = 15 × 8 or x ÷ 8 = 15
Step-by-step explanation:
- Total number of available tables = 15
- The number of guests who are seated at each of the 15 available tables = 8
Let 'x' be the total number of guests who are seated at the tables.
If have to determine how many guests (x) are seated at the tables, all we need is to multiply the 15 with 8.
so the equation becomes
x = 15 × 8 or x ÷ 8 = 15
Therefore, the equation that models this situation is:
x = 15 × 8 or x ÷ 8 = 15
Hello!
As you can see, all of these numbers just go down to the tenths, not the hundreds, so we cannot round.
Therefore, the numbers will remain the same.
I hope this helps!
