Puzzle for May 17, 2022  ( )

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

eq.1) A + B + C + D + E + F = 20 eq.2) F = B + D eq.3) D = C + E eq.4) A = C + D eq.5) E + F = A + C eq.6) D + E + F = A + 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 E + F with A + C (from eq.5): D + A + C = A + B + C Subtract A and C from both sides of the above equation: D + A + C – A – C = A + B + C – A – C which simplifies to B = D


  

Hint #2


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


  

Hint #3


Subtract C from each side of eq.3: D – C = C + E – C which becomes eq.3a) D – C = E


  

Hint #4


In eq.5, substitute D – C for E (eq.3a), 2×D for F, and C + D for A (from eq.4): D – C + 2×D = C + D + C which becomes 3×D – C = 2×C + D In the above equation, add C to both sides, and subtract D from both sides: 3×D – C + C – D = 2×C + D + C – D which makes 2×D = 3×C Divide both sides by 3: 2×D ÷ 3 = 3×C ÷ 3 which makes ⅔×D = C


  

Hint #5


Substitute ⅔×D for C in eq.4: A = ⅔×D + D which makes A = 1⅔×D


  

Hint #6


Substitute ⅔×D for C in eq.3a: D – ⅔×D = E which makes ⅓×D = E


  

Solution

Substitute 1⅔×D for A, D for B, ⅔×D for C, ⅓×D for E, and 2×D for F in eq.1: 1⅔×D + D + ⅔×D + D + ⅓×D + 2×D = 20 which simplifies to 6⅔×D = 20 Divide both sides of the above equation by 6⅔: 6⅔×D ÷ 6⅔ = 20 ÷ 6⅔ which means D = 3 making A = 1⅔×D = 1⅔×3 = 5 B = D = 3 C = ⅔×D = ⅔×3 = 2 E = ⅓×D = ⅓×3 = 1 F = 2×D = 2×3 = 6 and ABCDEF = 532316