LinkedIn Tango puzzle #743 for October 20, 2026: spoiler-free hints first, then the complete solution when you want it.
This is an earlier Tango puzzle. Latest Tango Answer: #767 (Nov 13)
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Reveal Hint #1Hint #1
Two moons at the end of row 3 force a sun at R3C4. Two moons at the start of row 4 force a sun at R4C3.
Reveal Hint #2Hint #2
Use the equality between R2C3 and R3C3, plus the three-suns-per-row rule. In the bottom row, equal neighbours R6C3 and R6C4 must both be suns.
Reveal Hint #3Hint #3
The first row is M M S S M S. The last row is M M S S M S too. Complete the middle rows by balancing each column.
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Tango #743 Solution
Row 1: Moon Moon Sun Sun Moon Sun. Row 2: Sun Sun Moon Moon Sun Moon. Row 3: Sun Sun Moon Sun Moon Moon. Row 4: Moon Moon Sun Moon Sun Sun. Row 5: Sun Sun Moon Moon Sun Moon. Row 6: Moon Moon Sun Sun Moon Sun.
How to Solve Tango #743
S = sun; M = moon. The two consecutive moons at R3C5–R3C6 force R3C4 to be a sun. Likewise, the moons at R4C1–R4C2 force R4C3 to be a sun.
- Row 1: M M S S M S.
- Row 2: S S M M S M.
- Row 3: S S M S M M.
- Row 4: M M S M S S.
- Row 5: S S M M S M.
- Row 6: M M S S M S.
The equal pairs are R1C3–R1C4 (suns), R2C3–R3C3 (moons), R4C4–R5C4 (moons) and R6C3–R6C4 (suns). All eight given symbols are preserved. Each row and column contains three suns and three moons, with no three identical symbols consecutively.
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