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$$a^7 \equiv a \pmod {42}.$$ There is no use telling you all what and how much I tried because I cannot even understand the problem itself left alone attempting it.

Nov 27, 2025 · By the Chinese remainder theorem, $x^n\equiv x\pmod n$ if and only if the congruences $x^n\equiv x\pmod {p_j^ {r_j}}$ are simultaneously true for $1\le j\le m$.

Aug 2, 2025 · Describe all integers $a$ such that $𝑎^ {111}\equiv 1\pmod {1111}$ I know that this means when you raise $𝑎$ to the power of $111$ and divide by $1111$, the remainder is $1$.

Prove that $ (\mathbb {Z}_n , +)$, the integers $\pmod {n}$ under addition, is a group. To show that this is a group, I know I need to show three things (in our text, we do not need to show …

Nov 20, 2020 · \pmod{m} will produce the parenthetical version of the mod operator (you don’t need to add parentheses); \bmod will produce the “mod” operator. \mod is the worst of the …

Jun 28, 2022 · Let $a$ and $b$ be integers and $m$ be a positive integer. Prove that $ab \equiv [ (a\pmod m)\cdot (b\pmod m)] \pmod m$ [duplicate] Ask Question Asked 3 years, 5 months …

Jul 4, 2015 · I'm currently taking Calculus I, so I'm new with mathematical notation. I always thought x mod 3 means the remainder when you divide x by 3. Am I wrong in thinking this is …

Aug 8, 2013 · What integers have order $6 \pmod {31}$? Ask Question Asked 12 years, 4 months ago Modified 12 years, 4 months ago

Feb 22, 2020 · This embodies a fact we know about congruences: $ka \cong kb \pmod {kc}$ if and only if $a \cong b \pmod {c}$. So the assumption is necessary to prevent the modulus …