Question

A. The following concept is very useful in Math 4000. Definition: Let $m \in \mathbb{N}$, and let $[a] \in \mathbb{Z}_m$. We say that $[b] \in \mathbb{Z}_m$ is the inverse of $[a]$ (mod $m$) when $[a][b] = [1]$ in $\mathbb{Z}_m$. In other words, $ab \equiv 1 \pmod{m}$, so multiplying $b$ "cancels" $a$. (It turns out $[b]$ is unique.) SAMPLE: In $\mathbb{Z}_7$, we have $[2]^{-1} = [4]$ because $2 \cdot 4 = 8 \equiv 1 \pmod{7}$. (a) Determine the inverse $[5]^{-1}$ in $\mathbb{Z}_{11}$, and briefly check your answer. NOTE: In scratchwork, you may experiment with all values of $[b] \in \mathbb{Z}_{11}$ to find your answer. However, when writing your final HW answer, you don't have to explain how you found the answer originally. (b) I'll give you the following inverse for free: $[77]^{-1} = [13]$ in $\mathbb{Z}_{100}$ (check this if you want) Use this to find the solution $[x] \in \mathbb{Z}_{100}$ to the following linear congruence: $[77][x] + [5] = [3]$ in $\mathbb{Z}_{100}$ Show your work, and simplify your answer to get $[x]$ where $0 \le x \le 99$. HINT: To "divide" the equation by $[77]$, what do we multiply to both sides instead?

          A. The following concept is very useful in Math 4000.
Definition: Let $m \in \mathbb{N}$, and let $[a] \in \mathbb{Z}_m$.
We say that $[b] \in \mathbb{Z}_m$ is the inverse of $[a]$ (mod $m$) when $[a][b] = [1]$ in $\mathbb{Z}_m$.
In other words, $ab \equiv 1 \pmod{m}$, so multiplying $b$ "cancels" $a$. (It turns out $[b]$ is
unique.)
SAMPLE: In $\mathbb{Z}_7$, we have $[2]^{-1} = [4]$ because $2 \cdot 4 = 8 \equiv 1 \pmod{7}$.
(a) Determine the inverse $[5]^{-1}$ in $\mathbb{Z}_{11}$, and briefly check your answer.
NOTE: In scratchwork, you may experiment with all values of $[b] \in \mathbb{Z}_{11}$ to find
your answer. However, when writing your final HW answer, you don't have to
explain how you found the answer originally.
(b) I'll give you the following inverse for free:
$[77]^{-1} = [13]$ in $\mathbb{Z}_{100}$ (check this if you want)
Use this to find the solution $[x] \in \mathbb{Z}_{100}$ to the following linear congruence:
$[77][x] + [5] = [3]$ in $\mathbb{Z}_{100}$
Show your work, and simplify your answer to get $[x]$ where $0 \le x \le 99$.
HINT: To "divide" the equation by $[77]$, what do we multiply to both sides instead?

A. The following concept is very useful in Math 4000.
Definition: Let m ∈ℕ, and let [a] ∈ℤm.
We say that [b] ∈ℤm is the inverse of [a] (mod m) when [a][b] = [1] in ℤm.
In other words, ab ≡ 1 m, so multiplying b "cancels" a. (It turns out [b] is
unique.)
SAMPLE: In ℤ7, we have [2]^-1 = [4] because 2 · 4 = 8 ≡ 1 7.
(a) Determine the inverse [5]^-1 in ℤ11, and briefly check your answer.
NOTE: In scratchwork, you may experiment with all values of [b] ∈ℤ11 to find
your answer. However, when writing your final HW answer, you don't have to
explain how you found the answer originally.
(b) I'll give you the following inverse for free:
[77]^-1 = [13] in ℤ100 (check this if you want)
Use this to find the solution [x] ∈ℤ100 to the following linear congruence:
[77][x] + [5] = [3] in ℤ100
Show your work, and simplify your answer to get [x] where 0 ≤ x ≤ 99.
HINT: To "divide" the equation by [77], what do we multiply to both sides instead?

Added by Crystal R.

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