Puzzle for May 26, 2022  ( )

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Find the 6-digit number ABCDEF by solving the following equations:

eq.1) D = A + C eq.2) C – D = A – C eq.3) E – C = C + D – E eq.4) F = A × B eq.5) A – E = E – (A × D) eq.6)* C ^ F = E mod A

A, B, C, D, E, and F each represent a one-digit non-negative integer.
* "C ^ F" means "C raised to the power of F". "E mod A" equals the remainder of (E ÷ A).
 
Many thanks to Judah S (age 15) for sending us this puzzle! Thank you, Judah!

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Hint #1


In eq.2, substitute (A + C) for D (from eq.1): C – (A + C) = A – C which becomes C – A – C = A – C which becomes –A = A – C Add A and C to both sides of the above equation: –A + A + C = A – C + A + C which makes C = 2×A


  

Hint #2


In eq.1, replace C with 2×A: D = A + 2×A which makes D = 3×A


  

Hint #3


In eq.3, replace C with 2×A, and D with 3×A: E – 2×A = 2×A + 3×A – E which becomes E – 2×A = 5×A – E Add 2×A and E to both sides of the equation above: E – 2×A + 2×A + E = 5×A – E + 2×A + E which becomes 2×E = 7×A Divide both sides by 2: 2×E ÷ 2 = 7×A ÷ 2 which makes E = 3½×A


  

Hint #4


In eq.5, substitute 3½×A for E, and 3×A for D: A – 3½×A = 3½×A – (A × 3×A) which becomes –2½×A = 3½×A – (A × 3×A) Add 2½×A and (A × 3×A) to both sides of the above equation: –2½×A + 2½×A + (A × 3×A) = 3½×A – (A × 3×A) + 2½×A + (A × 3×A) which simplifies to A × 3×A = 6×A Since A ≠ 0 (from eq.6), divide both sides of the above equation by 3×A: (A × 3×A) ÷ 3×A = 6×A ÷ 3×A which makes A = 2 and makes C = 2×A = 2 × 2 = 4 D = 3×A = 3 × 2 = 6 E = 3½×A = 3½ × 2 = 7


  

Hint #5


Substitute 4 for C, 7 for E, and 2 for A in eq.6: 4 ^ F = 7 mod 2 which means 4 ^ F = remainder of (7 ÷ 2) which makes 4 ^ F = 1 The above equation makes: F = 0


  

Solution

Substitute 0 for F, and 2 for A in eq.4: 0 = 2 × B which means 0 = B and makes ABCDEF = 204670