Puzzle for February 16, 2021  ( )

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

eq.1) A + B + C + D + E + F = 31 eq.2) B + D = A + F eq.3) C + D = E eq.4) D + F = C eq.5) E + F = A + B eq.6) F = A + D

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

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Hint #1


In eq.2, replace F with A + D (from eq.6): B + D = A + A + D which becomes B + D = 2×A + D Subtract D from each side of the above equation: B + D – D = 2×A + D – D which makes B = 2×A


  

Hint #2


Subtract C from both sides of eq.3: C + D – C = E – C which becomes D = E – C In eq.4, replace D with E – C: E – C + F = C Add C to both sides of the above equation: E – C + F + C = C + C which becomes eq.4a) E + F = 2×C


  

Hint #3


In eq.5, replace E + F with 2×C (from eq.4a), and B with 2×A: 2×C = A + 2×A which becomes 2×C = 3×A Divide both sides of the equation above by 2: 2×C ÷ 2 = 3×A ÷ 2 which makes C = 1½×A


  

Hint #4


In eq.4, substitute A + D for F (from eq.6), and 1½×A for C: D + A + D = 1½×A which becomes A + 2×D = 1½×A Subtract A from both sides of the above equation: A + 2×D – A = 1½×A – A which becomes 2×D = ½×A Divide both sides by 2: 2×D ÷ 2 = ½×A ÷ 2 which makes D = ¼×A


  

Hint #5


Substitute ¼×A for D in eq.6: F = A + ¼×A which makes F = 1¼×A


  

Hint #6


Substitute 1½×A for C, and ¼×A for D in eq.3: 1½×A + ¼×A = E which makes 1¾×A = E


  

Solution

Substitute 2×A for B, 1½×A for C, ¼×A for D, 1¾×A for E, and 1¼×A for F in eq.1: A + 2×A + 1½×A + ¼×A + 1¾×A + 1¼×A = 31 which simplifies to 7¾×A = 31 Divide both sides of the above equation by 7¾: 7¾×A ÷ 7¾ = 31 ÷ 7¾ which means A = 4 making B = 2×A = 2 × 4 = 8 C = 1½×A = 1½ × 4 = 6 D = ¼×A = ¼ × 4 = 1 E = 1¾×A = 1¾ × 4 = 7 F = 1¼×A = 1¼ × 4 = 5 and ABCDEF = 486175