Puzzle for October 28, 2021  ( )

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

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

A, B, C, D, E, and F each represent a one-digit non-negative integer.
*  "A mod C" equals the remainder of A ÷ C.

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Hint #1


In eq.4, substitute (B × C) for F (from eq.5): C = B ÷ (B × C) which is equivalent to C = (B ÷ B) × (1 ÷ C) which becomes C = 1 × (1 ÷ C) which means C = 1 ÷ C Multiply both sides of the above equation by C: C × C = (1 ÷ C) × C which becomes C² = 1 Since C is non-negative and C ≠ 0 (from eq.6), the above equation makes: C = 1


  

Hint #2


In eq.5, replace C with 1: F = B × 1 which means F = B


  

Hint #3


In eq.6, replace C with 1: E = A mod 1 which means E = remainder of (A ÷ 1) which means E = 0


  

Hint #4


In eq.1, substitute 0 for E, and 1 for C: A + 0 = 1 + F – 0 which makes eq.1a) A = 1 + F


  

Hint #5


Substitute F for B, 1 for C, 1 + F for A (from eq.1a), and 0 for E in eq.3: F + F – 1 = 1 + F + 1 + D – 0 which becomes 2×F – 1 = 2 + F + D Subtract 2 and F from each side of the above equation: 2×F – 1 – 2 – F = 2 + F + D – 2 – F which becomes eq.3a) F – 3 = D


  

Solution

Substitute (F – 3) for D (from eq.3a), 0 for E, 1 + F for A (from eq.1a), F for B, and 1 for C in eq.2: (F – 3) + 0 = 1 + F – F + 1 – (F – 3) + F which becomes F – 3 = 2 – F + 3 + F which becomes F – 3 = 5 Add 3 to both sides of the above equation: F – 3 + 3 = 5 + 3 which makes F = 8 making A = 1 + F = 1 + 8 = 9 (from eq.1a) B = F = 8 D = F – 3 = 8 – 3 = 5 (from eq.3a) and ABCDEF = 981508