Puzzle for February 25, 2022  ( )

Scratchpad

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

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

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

Scratchpad

 

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Hint #1


eq.4 may be written as: E = (D + F) ÷ 2 Multiply both sides of the equation above by 2: 2 × E = 2 × (D + F) ÷ 2 which becomes eq.4a) 2×E = D + F   eq.3 may be written as: eq.3a) C ÷ E = D + F – E


  

Hint #2


In eq.3a, replace D + F with 2×E (from eq.4a): C ÷ E = 2×E – E which makes C ÷ E = E Multiply both sides of the above equation by E: (C ÷ E) × E = E × E which makes C = E²


  

Hint #3


In eq.6, replace C with E²: F = E² mod E Since E² ÷ E = E then remainder of (E² ÷ E) = 0 making F = remainder of (E² ÷ E) = 0


  

Hint #4


In eq.4a, substitute 0 for F: 2×E = D + 0 which makes 2×E = D


  

Hint #5


Substitute E² for C, and 0 for F in eq.1: E² + 0 = B + E – 0 which becomes E² = B + E Subtract E from each side of the equation above: E² – E = B + E – E which makes eq.1a) E² – E = B


  

Hint #6


eq.5 may be written as: D = (B + C + E) ÷ 3 Multiply both sides of the above equation by 3: 3 × D = 3 × (B + C + E) ÷ 3 which becomes eq.5a) 3×D = B + C + E


  

Hint #7


Substitute (2×E) for D, E² – E for B (from eq.1a), and E² for C in eq.5a: 3×(2×E) = E² – E + E² + E which becomes 6×E = 2×E² Since E ≠ 0 (from eq.2 and eq.3), divide both sides of the above equation by 2×E: 6×E ÷ 2×E = 2×E² ÷ 2×E which makes 3 = E making B = E² – E = 3² – 3 = 9 – 3 = 6 (from eq.1a) C = E² = 3² = 9 D = 2×E = 2×3 = 6


  

Solution

Substitute 6 for B and D, 3 for E, and 0 for F in eq.2: 6 ÷ 3 = A – 6 + 0 which becomes 2 = A – 6 Add 6 to both sides of the equation above: 2 + 6 = A – 6 + 6 which makes 8 = A and makes ABCDEF = 869630