ExampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-07-31

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The system $x\equiv2\pmod3$ , $x\equiv3\pmod5$ , $x\equiv2\pmod7$ has the unique solution $x\equiv23\pmod{105}$

Example

The system

$x\equiv2\pmod3,\qquad x\equiv3\pmod5,\qquad x\equiv2\pmod7$

has exactly the solutions $x\equiv23\pmod{105}$ .

Facts & Assumptions

Given: The three displayed congruences.

[L1]

For a finite pairwise-coprime list of positive moduli, prescribed residues determine one class modulo their product (Chinese remainder theorem for a finite pairwise-coprime list: simultaneous residues determine one class modulo the product, and the resulting bijection preserves addition and multiplication).

[L2]

The relation $u\equiv v\pmod n$ means $n\mid(u-v)$ (Congruence modulo an integer: $a\equiv b\pmod n$ when $n\mid(a-b)$ , including the moduli $0$ and $1$ ).

Verification

technique · constructive

1.1

The numbers $3,5,7$ are pairwise coprime and have product $105$ . The complementary products $35,21,15$ satisfy $2\cdot35\equiv1\pmod3$ , $21\equiv1\pmod5$ , and $15\equiv1\pmod7$ , while each is divisible by the other two moduli.

2.1

Therefore $x:=2(2\cdot35)+3(21)+2(15)=233$ has the prescribed three residues, and $233=23+2\cdot105$ gives $x\equiv23\pmod{105}$ .

constructstep 1.1L2

3.1

By [L1], all simultaneous solutions form one class modulo $3\cdot5\cdot7=105$ ; since step 2.1 exhibits $23$ in that class, it is exactly $[23]_{105}$ .

step 2.1L1discharge-construct∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 47 results over 17 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

K. Conrad, The Chinese Remainder Theorem (standard reference, not scraped)
University of Southampton, Simultaneous Linear Congruences (standard reference, not scraped)

The system x≡2(mod3)x\equiv2\pmod3x≡2(mod3), x≡3(mod5)x\equiv3\pmod5x≡3(mod5), x≡2(mod7)x\equiv2\pmod7x≡2(mod7) has the unique solution x≡23(mod105)x\equiv23\pmod{105}x≡23(mod105)