Puzzle for September 1, 2022  ( )

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

eq.1) A + B + C + D + E + F = 12 eq.2) F – D = B + D eq.3) A – F = B + F eq.4) F = average (A, B, D) eq.5) B = average (C, E, F)

A, B, C, D, E, and F each represent a one-digit non-negative integer.

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Hint #1


Add D to both sides of eq.2: F – D + D = B + D + D which becomes eq.2a) F = B + 2×D   Add F to both sides of eq.3: A – F + F = B + F + F which becomes eq.3a) A = B + 2×F


  

Hint #2


In eq.3a, substitute (B + 2×D) for F (from eq.2a): A = B + 2×(B + 2×D) which becomes A = B + 2×B + 4×D which becomes eq.3b) A = 3×B + 4×D


  

Hint #3


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


  

Hint #4


In eq.4a, substitute (B + 2×D) for F (from eq.2a), and 3×B + 4×D for A (from eq.3b): 3×(B + 2×D) = 3×B + 4×D + B + D which becomes 3×B + 6×D = 4×B + 5×D Subtract 3×B and 5×D from each side of the equation above: 3×B + 6×D – 3×B – 5×D = 4×B + 5×D – 3×B – 5×D which simplifies to D = B


  

Hint #5


In eq.3b, replace D with B: A = 3×B + 4×B which makes A = 7×B


  

Hint #6


In eq.2a, replace D with B: F = B + 2×B which makes F = 3×B


  

Hint #7


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


  

Hint #8


Substitute 3×B for F in eq.5a: 3×B = C + E + 3×B Subtract 3×B from each side of the above equation: 3×B – 3×B = C + E + 3×B – 3×B which becomes 0 = C + E Since C and E must be non-negative, the above equation makes: C = 0 and E = 0


  

Solution

Substitute 7×B for A, 0 for C and E, B for D, and 3×B for F in eq.1: 7×B + B + 0 + B + 0 + 3×B = 12 which simplifies to 12×B = 12 Divide both sides of the above equation by 12: 12×B ÷ 12 = 12 ÷ 12 which means B = 1 making A = 7×B = 7 × 1 = 7 D = B = 1 F = 3×B = 3 × 1 = 3 and ABCDEF = 710103