What is the smallest integer $n$, greater than $1$, such that $n^{-1}\pmod{130}$ and $n^{-1}\pmod{231}$ are both defined?
1 answer:
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.
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Answer:
For first part, t= 0.18 sec or 3.4 sec.
For second part, t= 3.7 sec.
Step by step answer:
First part:
Set h(t) = 17and solve for tt.
-16t²+ 58t + 7= 17
-16t² + 58t - 10 = 0
Solve this quadratic equation for t. You should get 2 positive solutions. The lower value is the time to reach 17 on the way up, and the higher value is the time to reach 17 again, on the way down.
Second part:
Set h(t) = 0 and solve the resulting quadratic equation for t. You should get a negative solution (which you can discard), and a positive solution. The latter is your answer.
