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MATH / Given m≥2, denote by b^(−1) the inverse of bpmod{m}. That is, b^(−1) is the residue for which bb^{-1}equiv 1pmod{m}. Sadie wonders if (a+b)^(−1) is always congruent to …

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problem

Given m≥2m\geq 2, denote by b−1b^{-1} the inverse of $b\pmod{m}$. That is, b−1b^{-1} is the residue for which $bb^{-1}\equiv 1\pmod{m}$. Sadie wonders if (a+b)−1(a+b)^{-1} is always congruent to a−1+b−1a^{-1}+b^{-1} (modulo mm). She tries the example a=2a=2, b=3b=3, and m=7m=7. Let LL be the residue of $(2+3)^{-1}\pmod{7}$, and let RR be the residue of $2^{-1}+3^{-1}\pmod{7}$, where LL and RR are integers from 00 to 66 (inclusive). Find L−RL-R.
Plain-text mathematical notation (without MathML)
Given m≥2, denote by b^(−1) the inverse of $b\pmod{m}$. That is, b^(−1) is the residue for which $bb^{-1}\equiv 1\pmod{m}$. Sadie wonders if (a+b)^(−1) is always congruent to a^(−1)+b^(−1) (modulo m). She tries the example a=2, b=3, and m=7. Let L be the residue of $(2+3)^{-1}\pmod{7}$, and let R be the residue of $2^{-1}+3^{-1}\pmod{7}$, where L and R are integers from 0 to 6 (inclusive). Find L−R.
Original LaTeX notation
Given $m\geq 2$, denote by $b^{-1}$ the inverse of $b\pmod{m}$. That is, $b^{-1}$ is the residue for which $bb^{-1}\equiv 1\pmod{m}$. Sadie wonders if $(a+b)^{-1}$ is always congruent to $a^{-1}+b^{-1}$ (modulo $m$). She tries the example $a=2$, $b=3$, and $m=7$. Let $L$ be the residue of $(2+3)^{-1}\pmod{7}$, and let $R$ be the residue of $2^{-1}+3^{-1}\pmod{7}$, where $L$ and $R$ are integers from $0$ to $6$ (inclusive). Find $L-R$.

level

Level 5

type

Number Theory

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Source and history

task_id
464ae228-ed12-5b93-8bf9-756776ce8526
task_key
number~5ftheory--test--464ae228-ed12-5b93-8bf9-756776ce8526
task_revision_id
1
config / split
number_theory / test

Official source

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