Puzzle for May 25, 2021  ( )

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

eq.1) A + B + C + D + E + F = 33 eq.2) D = A + B eq.3) C = B + E eq.4) B + C = A + D eq.5) A + F = B + D eq.6) B + F = A + E

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

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Hint #1


In eq.4, replace D with A + B (from eq.2): B + C = A + A + B which becomes B + C = 2×A + B Subtract B from both sides of the equation above: B + C – B = 2×A + B – B which makes C = 2×A


  

Hint #2


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


  

Hint #3


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


  

Hint #4


Substitute 2×A for C, and 3×B – A for E (from eq.6a) in eq.3: 2×A = B + 3×B – A which becomes 2×A = 4×B – A Add A to both sides of the above equation: 2×A + A = 4×B – A + A which makes 3×A = 4×B Divide both sides by 4: 3×A ÷ 4 = 4×B ÷ 4 which makes ¾×A = B


  

Hint #5


Substitute ¾×A for B in eq.2: D = A + ¾×A which makes D = 1¾×A


  

Hint #6


Substitute (¾×A) for B in eq.5a: F = 2×(¾×A) which makes F = 1½×A


  

Hint #7


Substitute (¾×A) for B in eq.6a: 3×(¾×A) – A = E which becomes 2¼×A – A = E which makes 1¼×A = E


  

Solution

Substitute ¾×A for B, 2×A for C, 1¾×A for D, 1¼×A for E, and 1½×A for F in eq.1: A + ¾×A + 2×A + 1¾×A + 1¼×A + 1½×A = 33 which simplifies to 8¼×A = 33 Divide both sides of the above equation by 8¼: 8¼×A ÷ 8¼ = 33 ÷ 8¼ which means A = 4 making B = ¾×A = ¾ × 4 = 3 C = 2×A = 2 × 4 = 8 D = 1¾×A = 1¾ × 4 = 7 E = 1¼×A = 1¼ × 4 = 5 F = 1½×A = 1½ × 4 = 6 and ABCDEF = 438756