Puzzle for July 2, 2022  ( )

Scratchpad

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

eq.1) F = B + C eq.2) A + B = F – B eq.3) B + D = E ÷ D eq.4) A + D = (C × D) – E eq.5) E – B = B + (D × D) eq.6) C × E = (A × F) – E

A, B, C, D, E, and F each represent a one-digit positive integer.
 
Once again, we thank frequent contributor Judah S (age 15) for sending us an interesting puzzle. Thank you, Judah!

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Hint #1


Multiply both sides of eq.3 by D: D × (B + D) = D × (E ÷ D) which becomes D × (B + D) = E which may be written as eq.3a) (D × B) + (D × D) = E


  

Hint #2


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


  

Hint #3


In eq.3a, replace D with 2: (2 × B) + (2 × 2) = E which becomes eq.3b) 2×B + 4 = E


  

Hint #4


In eq.2, replace F with B + C (from eq.1): A + B = B + C – B which becomes eq.2a) A + B = C


  

Hint #5


In eq.4, substitute 2 for D, (A + B) for C (from eq.2a), and (2×B + 4) for E (from eq.3b): A + 2 = ((A + B) × 2) – (2×B + 4) which becomes A + 2 = 2×A + 2×B – 2×B – 4 which becomes A + 2 = 2×A – 4 In the equation above, subtract A from both sides, and add 4 to both sides: A + 2 – A + 4 = 2×A – 4 – A + 4 which makes 6 = A


  

Hint #6


Substitute 6 for A in eq.2a: eq.2b) 6 + B = C


  

Hint #7


Substitute 6 for A in eq.2: 6 + B = F – B Add B to both sides of the above equation: 6 + B + B = F – B + B which becomes eq.2c) 6 + 2×B = F


  

Hint #8


In eq.6, substitute (6 + B) for C (from eq.2b), (2×B + 4) for E (from eq.3b), 6 for A, and (6 + 2×B) for F (from eq.2c) in eq.6: (6 + B) × (2×B + 4) = (6 × (6 + 2×B)) – (2×B + 4) which becomes 12×B + 24 + 2×B² + 4×B = 36 + 12×B – 2×B – 4 which becomes 16×B + 24 + 2×B² = 32 + 10×B Subtract 32 and 10×B from each side of the equation above: 16×B + 24 + 2×B² – 32 – 10×B = 32 + 10×B – 32 – 10×B which becomes 6×B – 8 + 2×B² = 0 Divide both sides by 2: (6×B – 8 + 2×B²) ÷ 2 = 0 ÷ 2 which becomes 3×B – 4 + B² = 0 which may be written as the product of two terms eq.6a) (B + 4) × (B – 1) = 0


  

Hint #9


To make eq.6a true, either: B = –4 or B = 1 Since B must be positive, then B ≠ –4 which means B = 1


  

Solution

Since B = 1, then: C = 6 + B = 6 + 1 = 7 (from eq.2b) E = 2×B + 4 = 2×1 + 4 = 2 + 4 = 6 (from eq.3b) F = 6 + 2×B = 6 + 2×1 = 6 + 2 = 8 (from eq.2c) and ABCDEF = 617268