85% will be correct i feel like
Yes, 23 has an inverse mod 1000 because gcd(23, 1000) = 1 (i.e. they are coprime).
Let <em>x</em> be the inverse. Then <em>x</em> is such that
23<em>x</em> ≡ 1 (mod 1000)
Use the Euclidean algorithm to solve for <em>x</em> :
1000 = 43×23 + 11
23 = 2×11 + 1
→ 1 ≡ 23 - 2×11 (mod 1000)
→ 1 ≡ 23 - 2×(1000 - 43×23) (mod 1000)
→ 1 ≡ 23 - 2×1000 + 86×23 (mod 1000)
→ 1 ≡ 87×23 - 2×1000 ≡ 87×23 (mod 1000)
→ 23⁻¹ ≡ 87 (mod 1000)
Answer:
I know
Step-by-step explanation:
Answer:
1. Proved down
2. proved down
3. f(10) = -20 - 5 - 5 - 5 - 5 - 5 - 5 - 5 - 5 - 5 - 5
Step-by-step explanation:
Let us explain how to solve the question
∵ f(0) = -20, f(n) = f(n - 1) - 5 for n > 1
→ That means we have an arithmetic sequence with constant
difference -5 and first term -20
1. → f(1) means we need to find the second term, which equal the
term - 5
∵ f(1) means n = 1
∴ f(1) = f(1 - 1) - 5
∴ f(1) = f(0) - 5
∵ f(0) = -20
∴ f(1) = -20 - 5 → Proved
2. → f(3) means we need to find the third term, which equal the
second term - 5
∵ f(3) means n = 3
∴ f(3) = f(3 - 1) - 5
∴ f(3) = f(2) - 5
→ f(2) = f(1) - 5
∵ f(1) = -20 - 5
∴ f(2) = [-20 - 5] - 5 = -20 - 5 - 5
∴ f(3) = [-20 - 5 - 5] - 5
∴ f(3) = -20 - 5 - 5 - 5 → Proved
3. → From 1 and 2 we notice that the number of -5 is equal to n,
at n = 1 there is one (-5), when n= 3 there are three (-5)
∵ n = 10
∴ There are ten (-5)
∴ f(10) = -20 - 5(10)
∴ f(10) = -20 - 5 - 5 - 5 - 5 - 5 - 5 - 5 - 5 - 5 - 5 → Proved
