Puzzle for October 29, 2020  ( )

Scratchpad

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

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

A, B, C, D, E, and F each represent a one-digit non-negative integer.
*  EF is a 2-digit number (not E×F).

Scratchpad

 

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Hint #1


eq.5 may be written as: A = (D + F) ÷ 2 Multiply both sides of the above equation by 2: 2 × A = 2 × (D + F) ÷ 2 which becomes eq.5a) 2×A = D + F


  

Hint #2


In eq.5a, replace D with A + C (from eq.1): 2×A = A + C + F Subtract A and C from each side of the equation above: 2×A – A – C = A + C + F – A – C which becomes eq.5b) A – C = F


  

Hint #3


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


  

Hint #4


In eq.1, substitute A – 2×C for D (from eq.2a): A + C = A – 2×C In the equation above, subtract A from both sides, and add 2×C to both sides: A + C – A + 2×C = A – 2×C – A + 2×C which simplifies to 3×C = 0 which means C = 0


  

Hint #5


Substitute 0 for C in eq.1: D = A + 0 which makes D = A


  

Hint #6


Substitute 0 for C in eq.2: 0 + D = F which makes D = F and also means A = D = F


  

Hint #7


Substitute D for F in eq.4: D ÷ D = E which makes 1 = E


  

Hint #8


Substitute 1 for E in eq.3: B + 1 = A – B In the above equation, subtract 1 from both sides, and add B to both sides: B + 1 – 1 + B = A – B – 1 + B which becomes 2×B = A – 1 Divide both sides by 2: 2×B ÷ 2 = (A – 1) ÷ 2 which makes eq.3a) B = (A – 1) ÷ 2


  

Solution

eq.6 may be written as: 10×E + F = A + D + E Substitute 1 for E, and A for F and D in the above equation: 10×1 + A = A + A + 1 which becomes 10 + A = 2×A + 1 Subtract A and 1 from each side of the above equation: 2×A + 1 – A – 1 = 10 + A – A – 1 which makes A = 9 making B = (A – 1) ÷ 2 = (9 – 1) ÷ 2 = 8 ÷ 2 = 4 (from eq.3a) D = F = A = 9 and ABCDEF = 940919