Puzzle for February 2, 2020  ( )

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

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

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

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Hint #1


Add C to both sides of eq.4: C + F + C = B – C + D + C which becomes eq.4a) 2×C + F = B + D   Add F and D to both sides of eq.2: B – F + F + D = C – D + F + F + D which becomes eq.2a) B + D = C + 2×F


  

Hint #2


In eq.2a, replace B + D with 2×C + F (from eq.4a): 2×C + F = C + 2×F Subtract both F and C from each side of the equation above: 2×C + F – F – C = C + 2×F – F – C which makes C = F


  

Hint #3


In eq.3, replace F with C: D + C = C + E Subtract C from each side of the above equation: D + C – C = C + E – C which makes D = E


  

Hint #4


Substitute D for E, and C for F in eq.1: D + C = A which may be written as eq.1a) C + D = A


  

Hint #5


Substitute D for E in eq.5: B = D × D which may be written as B = D²


  

Hint #6


Substitute C for F in eq.6: A + B = C × C which may be written as eq.6a) A + B = C²


  

Hint #7


Substitute C + D for A (from eq.1a), and D² for B in eq.6a: C + D + D² = C² Subtract D² from each side of the equation above: C + D + D² – D² = C² – D² which becomes eq.6b) C + D = C² – D²


  

Hint #8


eq.6b may be written as: C + D = (C – D) × (C + D) Divide both sides of the equation above by (C + D): (C + D) ÷ (C + D) = (C – D) × (C + D) ÷ (C + D) which becomes 1 = C – D Add D to each side of the equation above: 1 + D = C – D + D which makes 1 + D = C and also makes eq.6c) F = C = 1 + D


  

Hint #9


Substitute D for E, and 1 + D for F (from eq.6c) in eq.1: A = D + 1 + D which makes eq.1a) A = 2×D + 1


  

Hint #10


Substitute 1 + D for C and F (from eq.6c) in eq.2a: B + D = 1 + D + 2×(1 + D) which is equivalent to B + D = 1 + D + 2 + 2×D which becomes B + D = 3×D + 3 Subtract D from both sides of the equation above: B + D – D = 3×D + 3 – D which makes eq.2b) B = 2×D + 3


  

Hint #11


Substitute 2×D + 1 for A (from eq.1a), 2×D + 3 for B (from eq.2b), and 1 + D for C (from eq.6c) in eq.6a: 2×D + 1 + 2×D + 3 = (1 + D)² which becomes 4×D + 4 = D² + 2×D + 1 Subtract both 4×D and 4 from each side of the above equation: 4×D + 4 – 4×D – 4 = D² + 2×D + 1 – 4×D – 4 which becomes eq.6d) 0 = D² – 2×D – 3


  

Solution

eq.6d is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for D in eq.6d yields: D = { (–1)×(–2) ± sq.rt.[(–2)² – (4 × (1) × (–3))] } ÷ (2 × (1)) which becomes D = {2 ± sq.rt.(4 – (–12))} ÷ 2 which becomes D = {2 ± sq.rt.(16)} ÷ 2 which becomes D = (2 ± 4) ÷ 2 In the above equation, either D = (2 + 4) ÷ 2 = 6 ÷ 2 = 3 or D = (2 – 4) ÷ 2 = –2 ÷ 2 = –1 Since D must be a positive integer, then D ≠ –1 and therefore makes D = 3 making A = 2×D + 1 = 2×3 + 1 = 6 + 1 = 7 (from eq.1a) B = 2×D + 3 = 2×3 + 3 = 6 + 3 = 9 (from eq.2b) F = C = 1 + D = 1 + 3 = 4 (from eq.6c) E = D = 3 and ABCDEF = 794334