Puzzle for November 14, 2020  ( )

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

eq.1) A + B = C – B eq.2) F – B = A + C eq.3) D – C = B + C – (F – D) eq.4) A + F = average (A, C, D, E) eq.5)* CD = E + F eq.6)** E mod F = A ÷ D

A, B, C, D, E, and F each represent a one-digit non-negative integer.
*  CD is a 2-digit number (not C×D).
**  "E mod F" equals the remainder of E ÷ F.

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Hint #1


eq.3 is equivalent to: D – C = B + C – F + D In the equation above, add C and F to both sides, and subtract D from both sides: D – C + C + F – D = B + C – F + D + C + F – D which simplifies to eq.3a) F = B + 2×C


  

Hint #2


In eq.2, replace F with B + 2×C (from eq.3a): B + 2×C – B = A + C which becomes 2×C = A + C Subtract C from both sides of the above equation: 2×C – C = A + C – C which makes C = A


  

Hint #3


In eq.1, replace C with A: A + B = A – B In the above equation, add B to both sides, and subtract A from both sides: A + B + B – A = A – B + B – A which simplifies to 2×B = 0 which means B = 0


  

Hint #4


In eq.3a, substitute 0 for B: F = 0 + 2×C which makes F = 2×C


  

Hint #5


eq.5 may be written as: 10×C + D = E + F Substitute 2×C for F in the above equation: 10×C + D = E + 2×C Subtract 2×C from each side: 10×C + D – 2×C = E + 2×C – 2×C which becomes eq.5a) 8×C + D = E


  

Hint #6


eq.4 may be written as: A + F = (A + C + D + E) ÷ 4 Multiply both sides of the above equation by 4: 4 × (A + F) = 4 × (A + C + D + E) ÷ 4 which becomes eq.4a) 4×A + 4×F = A + C + D + E


  

Hint #7


In eq.4a, substitute C for A, (2×C) for F, and 8×C + D for E (from eq.5a): 4×C + 4×(2×C) = C + C + D + 8×C + D which becomes 4×C + 8×C = 10×C + 2×D which becomes 12×C = 10×C + 2×D Subtract 10×C from each side of the equation above: 12×C – 10×C = 10×C + 2×D – 10×C which makes 2×C = 2×D Divide both sides by 2: 2×C ÷ 2 = 2×D ÷ 2 which means C = D


  

Hint #8


Substitute C for D in eq.5a: 8×C + C = E which makes 9×C = E


  

Solution

In eq.6, substitute 9×C for E, 2×C for F, C for A, and C for D: 9×C mod 2×C = C ÷ C which becomes eq.6a) 9×C mod 2×C = 1   A property of the mod operation is -- if both numbers (or terms) have a factor in common, it may be factored out. Mathematically, the above property is expressed as: (m×C) mod (n×C) = (m mod n) × C   Applying the above property to eq.6a yields: (9 mod 2) × C = 1 which becomes 1 × C = 1 which means C = 1 making A = C = 1 D = C = 1 E = 9×C = 9 × 1 = 9 F = 2×C = 2 × 1 = 2 and ABCDEF = 101192