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The Euclidean algorithm computes the greatest common divisor; back-substitution finds Bézout coefficients and a modular inverse.
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Formula
a, m, x, y are integers; residue class modulo m; no physical unit
| Symbol | Meaning | Unit |
|---|---|---|
| Integer a and positive modulus m | integer; m>0 | |
| Greatest common divisor | integer | |
| Bézout coefficients | integer | |
| a−1 mod m | Multiplicative inverse as residue class | modulo m |
\gcd(a,m)=\gcd(m,a\bmod m),\quad ax+my=\gcd(a,m)Repeat division with remainder until the remainder is zero; the last nonzero remainder is the gcd. Back-substitution yields x and y with ax+my=gcd(a,m). An inverse of a modulo m exists exactly when gcd(a,m)=1. Thus modular reciprocals do not exist for every a; for composite m, coprimality to m—not merely oddness—is required.
Repeat division with remainder until the remainder is zero; the last nonzero remainder is the gcd. Back-substitution yields x and y with ax+my=gcd(a,m). An inverse of a modulo m exists exactly when gcd(a,m)=1. Thus modular reciprocals do not exist for every a; for composite m, coprimality to m—not merely oddness—is required.
The Euclidean algorithm computes the greatest common divisor; back-substitution finds Bézout coefficients and a modular inverse.
Defining relation
The equation states the defining relation within its stated domain.
Use the example for Euclidean algorithm and modular inverse to explain which inputs determine the requested output.
The example connects concrete inputs to its result.
The Euclidean algorithm computes the greatest common divisor; back-substitution finds Bézout coefficients and a modular inverse.
The relation applies only under its stated assumptions.
Find the inverse of 35 modulo 12 using extended Euclid.
35=2·12+11; 12=1·11+1; hence 1=3·12−35 and 35−1≡−1≡11 (mod 12).
Check: 35·11=385=32·12+1.
Applying the relation to a different case without checking its stated assumptions.
Check the validity range before substituting values: Repeat division with remainder until the remainder is zero; the last nonzero remainder is the gcd. Back-substitution yields x and y with ax+my=gcd(a,m). An inverse of a modulo m exists.
Formula cluster
The Euclidean algorithm computes the greatest common divisor; back-substitution finds Bézout coefficients and a modular inverse.
gcd(35,12): 35=2·12+11, 12=1·11+1, 11=11·1. Back-substitution gives 1=12−11=3·12−35, so 35−1 mod 12=−1 mod 12=11; check 35·11=385≡1 (mod 12).
Compute modular division, congruences, number-theory exercises and RSA teaching examples.
Here is how to work through a typical Euclidean algorithm and modular inverse (gcd(a,m)=1 ⇔ a⁻¹ mod m existiert) task step by step:
Task
Use the example for Euclidean algorithm and modular inverse to explain which inputs determine the requested output.
Solution path
The example connects concrete inputs to its result.
The Euclidean algorithm computes the greatest common divisor; back-substitution finds Bézout coefficients and a modular inverse.
The relation applies only under its stated assumptions.
Task
Find the inverse of 35 modulo 12 using extended Euclid.
Solution path
35=2·12+11; 12=1·11+1; hence 1=3·12−35 and 35−1≡−1≡11 (mod 12).
Check: 35·11=385=32·12+1.
The relation \gcd(a,m)=\gcd(m,a\bmod m),\quad ax+my=\gcd(a,m) summarizes The Euclidean algorithm computes the greatest common divisor; back-substitution finds Bézout coefficients and a modular inverse. Read it as a model for defined quantities, not as an automatic claim about every real situation. In the example gcd(35,12): 35=2·12+11, 12=1·11+1, 11=11·1. Back-substitution gives 1=12−11=3·12−35, so 35−1 mod 12=−1 mod 12=11; check 35·11=385≡1 (mod 12)., symbols are connected to concrete values and the result is interpreted in context. Validity limit: Repeat division with remainder until the remainder is zero; the last nonzero remainder is the gcd. Back-substitution yields x and y with ax+my=gcd(a,m). An inverse of a modulo m exists Source: MIT OpenCourseWare, 6.042J Mathematics for Computer Science, complete course notes, modular arithmetic and multiplicative inverses.
First write down the requested quantity and all given values with their symbols. Substitute them into the defining relation and check unit consistency. The example gcd(35,12): 35=2·12+11, 12=1·11+1, 11=11·1. Back-substitution gives 1=12−11=3·12−35, so 35−1 mod 12=−1 mod 12=11; check 35·11=385≡1 (mod 12). provides a worked check: formula and result must have compatible scale and meaning; round only at the end. Validity limit: Repeat division with remainder until the remainder is zero; the last nonzero remainder is the gcd. Back-substitution yields x and y with ax+my=gcd(a,m). An inverse of a modulo m exists Source: MIT OpenCourseWare, 6.042J Mathematics for Computer Science, complete course notes, modular arithmetic and multiplicative inverses.
The formula applies only when quantities are defined as in the source and background. Check boundary cases, measurement conditions and whether an approximation rather than equality is intended. Substitution can be algebraically correct yet technically wrong when assumptions fail or the population or system boundary changes. Validity limit: Repeat division with remainder until the remainder is zero; the last nonzero remainder is the gcd. Back-substitution yields x and y with ax+my=gcd(a,m). An inverse of a modulo m exists Source: MIT OpenCourseWare, 6.042J Mathematics for Computer Science, complete course notes, modular arithmetic and multiplicative inverses.
The stated units are: a, m, x, y are integers; residue class modulo m; no physical unit. Convert values to a consistent unit system before substitution; counts, percentages and dimensionless ratios are not interchangeable. The variable list states the role of each symbol. Dimensional analysis catches many errors but does not replace checking the formula's assumptions. Validity limit: Repeat division with remainder until the remainder is zero; the last nonzero remainder is the gcd. Back-substitution yields x and y with ax+my=gcd(a,m). An inverse of a modulo m exists Source: MIT OpenCourseWare, 6.042J Mathematics for Computer Science, complete course notes, modular arithmetic and multiplicative inverses.
Comparison helps when both quantities address the same question but emphasize different properties. Report the definition, input data, threshold or averaging with the value. The source MIT OpenCourseWare, 6.042J Mathematics for Computer Science, complete course notes, modular arithmetic and multiplicative inverses. supports the definition used here; other conventions can differ. Choose the measure to match the learning or analysis goal. Validity limit: Repeat division with remainder until the remainder is zero; the last nonzero remainder is the gcd. Back-substitution yields x and y with ax+my=gcd(a,m). An inverse of a modulo m exists Source: MIT OpenCourseWare, 6.042J Mathematics for Computer Science, complete course notes, modular arithmetic and multiplicative inverses.
Quanta exam set
Curated exam set for "Euclidean algorithm and modular inverse":
Question (front)
Which formula describes Euclidean algorithm and modular inverse?
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Question (front)
How do you rearrange Euclidean algorithm and modular inverse for Defining relation?
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Question (front)
Which common mistake happens with Euclidean algorithm and modular inverse?
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Create a curated FSRS exam set for gcd(a,m)=1 ⇔ a⁻¹ mod m existiert: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.
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