Puzzle for September 6, 2020  ( )

Scratchpad

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

eq.1) A + B + C + D + E + F = 25 eq.2) D = B – D + F eq.3) A + C – F = D + F eq.4) B = A + D eq.5) E = A + C + F eq.6) E + F = (A × C) – F

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

Scratchpad

 

Help Area

Hint #1


In eq.2, replace B with A + D (from eq.4): D = A + D – D + F which becomes eq.2a) D = A + F


  

Hint #2


In eq.3, replace D with A + F (from eq.2a): A + C – F = A + F + F which becomes A + C – F = A + 2×F In the equation above, add F to each side, and subtract A from each side: A + C – F + F – A = A + 2×F + F – A which makes eq.3a) C = 3×F


  

Hint #3


In eq.5, substitute 3×F for C: E = A + 3×F + F which becomes eq.5a) E = A + 4×F


  

Hint #4


Substitute A + 4×F for E (from eq.5a), and 3×F for C in eq.6: A + 4×F + F = (A × 3×F) – F which becomes A + 5×F = (A × 3×F) – F Subtract 5×F from each side: A + 5×F – 5×F = (A × 3×F) – F – 5×F which becomes A = A × 3×F – 6×F which may be written as A = F × (3×A – 6) (implies A ≠ 2) Divide both sides by (3×A – 6): A ÷ (3×A – 6) = F × (3×A – 6) ÷ (3×A – 6) which becomes eq.6a) A ÷ (3×A – 6) = F


  

Hint #5


Substitute A + F for D (from eq.2a) in eq.4: B = A + A + F which becomes B = 2×A + F Substitute (A ÷ (3×A – 6)) for F (from eq.6a) in the above equation: eq.4a) B = 2×A + (A ÷ (3×A – 6))


  

Hint #6


Substitute (A ÷ (3×A – 6)) for F (from eq.6a) in eq.3a: eq.3b) C = 3×(A ÷ (3×A – 6))


  

Hint #7


Substitute (A ÷ (3×A – 6)) for F (from eq.6a) in eq.2a: eq.2b) D = A + (A ÷ (3×A – 6))


  

Hint #8


Substitute (A ÷ (3×A – 6)) for F (from eq.6a) in eq.5a: eq.5b) E = A + 4×(A ÷ (3×A – 6))


  

Hint #9


In eq.1, substitute -- 2×A + (A ÷ (3×A – 6)) for B (from eq.4a), 3×(A ÷ (3×A – 6)) for C (from eq.3b), A + (A ÷ (3×A – 6)) for D (from eq.2b), A + 4×(A ÷ (3×A – 6)) for E (from eq.5b), and A ÷ (3×A – 6) for F (from eq.6a): A + 2×A + (A ÷ (3×A – 6)) + 3×(A ÷ (3×A – 6)) + A + (A ÷ (3×A – 6)) + A + 4×(A ÷ (3×A – 6)) + A ÷ (3×A – 6) = 25 which simplifies to eq.1a) 5×A + 10×(A ÷ (3×A – 6)) = 25


  

Hint #10


Subtract 5×A from both sides of eq.1a: 5×A + 10×(A ÷ (3×A – 6)) – 5×A = 25 – 5×A which becomes 10×(A ÷ (3×A – 6)) = 25 – 5×A Multiply both sides of the above equation by (3×A – 6): 10×(A ÷ (3×A – 6)) × (3×A – 6) = (25 – 5×A) × (3×A – 6) which simplifies to 10×A = 75×A – 15×A² – 150 + 30×A which becomes 10×A = 105×A – 15×A² – 150 Subtract 10×A from each side: 10×A – 10×A = 105×A – 15×A² – 150 – 10×A which becomes 0 = 95×A – 15×A² – 150 Divide both sides by 5: 0 ÷ 5 = (95×A – 15×A² – 150) ÷ 5 which becomes 0 = 19×A – 3×A² – 30 which may be written as eq.1b) 0 = –3×A² + 19×A – 30


  

Solution

eq.1b is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for A in eq.1b yields: A = {–19 ± sq.rt.(19² – (4×(–3)×(–30)))} ÷ (2×(–3)) which becomes A = {–19 ± sq.rt.(361 – 360)} ÷ (–6) which becomes A = {–19 ± sq.rt.(1)} ÷ (–6) which becomes A = (–19 ± 1) ÷ (–6) In the above equation, either A = (–19 + 1) ÷ (–6) = –18 ÷ (–6) = 3 or A = (–19 – 1) ÷ (–6) = –20 ÷ (–6) = 3⅓ Since A must be an integer, then A ≠ 3⅓ and therefore makes A = 3 making B = 2×3 + (3 ÷ (3×3 – 6)) = 6 + (3 ÷ (9 – 6)) = 6 + (3 ÷ 3) = 6 + 1 = 7 (from eq.4a) C = 3×(3 ÷ (3×3 – 6)) = 3×(3 ÷ (9 – 6)) = 3×(3 ÷ 3) = 3×1 = 3 (from eq.3b) D = 3 + (3 ÷ (3×3 – 6)) = 3 + (3 ÷ (9 – 6)) = 3 + (3 ÷ 3) = 3 + 1 = 4 (from eq.2b) E = 3 + 4×(3 ÷ (3×3 – 6)) = 3 + 4×(3 ÷ (9 – 6)) = 3 + 4×(3 ÷ 3) = 3 + 4×1 = 7 (from eq.5b) F = 3 ÷ (3×3 – 6) = 3 ÷ (9 – 6) = 3 ÷ 3 = 1 (from eq.6a) and ABCDEF = 373471