ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-07-31

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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The congruence $14 x \equiv 30 (m o d 100)$ has exactly the two solutions $x \equiv 45, 95 (m o d 100)$

Example

The congruence $14 x \equiv 30 (m o d 100)$ has exactly the solution classes

$[45]_{100} and [95]_{100} .$

Facts & Assumptions

Given: The congruence $14 x \equiv 30 (m o d 100)$ .

[L1]

A linear congruence $a x \equiv b (m o d n)$ is solvable exactly when $\gcd (a, n) ∣ b$ , and then has exactly $\gcd (a, n)$ solution classes modulo $n$ (For $n \geq 1$ , $a x \equiv b (m o d n)$ is solvable exactly when $\gcd (a, n) ∣ b$ , and then has exactly $\gcd (a, n)$ solution classes modulo $n$ ).

[L2]

The extended Euclidean algorithm produces gcds and Bézout coefficients (The extended Euclidean algorithm: the same descent produces integers $x, y$ with $a x + b y = \gcd (a, b)$ , so Bézout coefficients are computed and not merely shown to exist).

[L3]

Congruence modulo $n$ means divisibility of a difference by $n$ (Congruence modulo an integer: $a \equiv b (m o d n)$ when $n ∣ (a - b)$ , including the moduli $0$ and $1$ ).

Verification

technique · direct

1.1L1L2L3

The Euclidean algorithm gives $\gcd (14, 100) = 2$ , and $2 ∣ 30$ . Cancelling the common factor reduces the congruence to $7 x \equiv 15 (m o d 50)$ .

1.2L3

Since $7 \cdot 43 = 301 \equiv 1 (m o d 50)$ , multiplication by $43$ gives $x \equiv 43 \cdot 15 = 645 \equiv 45 (m o d 50)$ .

1.3L3

The two lifts of this class modulo $100$ are $45$ and $45 + 50 = 95$ , and both verify the original congruence: $14 \cdot 45 - 30 = 600$ and $14 \cdot 95 - 30 = 1300$ are divisible by $100$ .

2.1step 1.1step 1.2step 1.3L1∎

By [L1] there are exactly two solution classes modulo $100$ , so the two distinct classes in step 1.3 exhaust them.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

21 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Mathematics LibreTexts, Linear Congruences (standard reference, not scraped)
UPC, Modular Arithmetic notes (standard reference, not scraped)

The congruence 14x≡30(mod100)14x≡30(mod100) has exactly the two solutions x≡45,95(mod100)x≡45,95(mod100)