Puzzle for March 23, 2021  ( )

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

eq.1) A + B + C + D + E + F = 25 eq.2) C + D = A eq.3) B + D = C eq.4) A = B + E eq.5) F = C + E eq.6) E + F = B + C

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

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Hint #1


In eq.6, replace F with C + E (from eq.5): E + C + E = B + C which becomes C + 2×E = B + C Subtract C from both sides of the above equation: C + 2×E – C = B + C – C which makes 2×E = B


  

Hint #2


In eq.4, replace B with 2×E: A = 2×E + E which makes A = 3×E


  

Hint #3


In eq.2, substitute 3×E for A: eq.2a) C + D = 3×E


  

Hint #4


In eq.3, substitute 2×E for B: eq.3a) 2×E + D = C


  

Hint #5


Substitute 2×E + D for C (from eq.3a) into eq.2a: 2×E + D + D = 3×E which becomes 2×E + 2×D = 3×E Subtract 2×E from each side of the equation above: 2×E + 2×D – 2×E = 3×E – 2×E which becomes 2×D = E Divide both sides by 2: 2×D ÷ 2 = E ÷ 2 which makes D = ½×E


  

Hint #6


Substitute 2×E for B, and ½×E for D in eq.3: 2×E + ½×E = C which makes 2½×E = C


  

Hint #7


Substitute 2½×E for C in eq.5: F = 2½×E + E which makes F = 3½×E


  

Solution

Substitute 3×E for A, 2×E for B, 2½×E for C, ½×E for D, and 3½×E for F in eq.1: 3×E + 2×E + 2½×E + ½×E + E + 3½×E = 25 which simplifies to 12½×E = 25 Divide both sides of the above equation by 12½: 12½×E ÷ 12½ = 25 ÷ 12½ which means E = 2 making A = 3×E = 3 × 2 = 6 B = 2×E = 2 × 2 = 4 C = 2½×E = 2½ × 2 = 5 D = ½×E = ½ × 2 = 1 F = 3½×E = 3½ × 2 = 7 and ABCDEF = 645127