Puzzle for August 27, 2022  ( )

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

eq.1) A + B + C + D + E + F = 31 eq.2) D + F = A + E eq.3) A – E = C + D eq.4) C + E = F – D – E eq.5) F ÷ C = A – B + C – F eq.6)* C ^ D = A – B

A, B, C, D, E, and F each represent a one-digit non-negative integer.
*  "C ^ D" means "C raised to the power of D".

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Hint #1


Add D and E to both sides of eq.4: C + E + D + E = F – D – E + D + E which becomes C + 2×E + D = F which may be written as eq.4a) C + D + 2×E = F


  

Hint #2


In eq.4a, replace C + D with A – E (from eq.3): A – E + 2×E = F which becomes eq.4b) A + E = F


  

Hint #3


In eq.2, substitute F for A + E (from eq.4b): D + F = F Subtract F from each side of the equation above: D + F – F = F – F which makes D = 0


  

Hint #4


Substitute 0 for D in eq.6: C ^ 0 = A – B which becomes 1 = A – B (assumes C ≠ 0) Add B to both sides of the above equation: 1 + B = A – B + B which becomes eq.6a) 1 + B = A


  

Hint #5


Substitute 1 + B for A (from eq.6a) into eq.5: F ÷ C = 1 + B – B + C – F which becomes F ÷ C = 1 + C – F Multiply both sides of the above equation by C: C × (F ÷ C) = C × (1 + C – F) which becomes F = (C × 1) + (C × C) – (C × F) which is the same as eq.5a) F = C + C² – (C × F)


  

Hint #6


Add (C × F) to both sides of eq.5a: F + (C × F) = C + C² – (C × F) + (C × F) which becomes F + (C × F) = C + C² which may be written as eq.5b) F × (1 + C) = C × (1 + C)


  

Hint #7


Since (1 + C) ≠ –0 (C ≠ –1, because C is a non-negative integer), divide both sides of eq.5b by (1 + C): F × (1 + C) ÷ (1 + C) = C × (1 + C) ÷ (1 + C) which makes F = C


  

Hint #8


Substitute F for C, and 0 for D in eq.4a: F + 0 + 2×E = F which becomes F + 2×E = F Subtract F from each side of the above equation: F + 2×E – F = F – F which becomes 2×E = 0 which means E = 0


  

Hint #9


Substitute 0 for E in eq.4b: A + 0 = F which makes A = F


  

Hint #10


Substitute F for A in eq.6a: 1 + B = F Subtract 1 from both sides of the above equation: 1 + B – 1 = F – 1 which becomes eq.6b) B = F – 1


  

Solution

Substitute F for A and C, F – 1 for B (from eq.6b), and 0 for D and E in eq.1: F + F – 1 + F + 0 + 0 + F = 31 which simplifies to 4×F – 1 = 31 Add 1 to both sides of the equation above: 4×F – 1 + 1 = 31 + 1 which makes 4×F = 32 Divide both sides by 4: 4×F ÷ 4 = 32 ÷ 4 which means F = 8 making A = C = F = 8 B = F – 1 = 8 – 1 = 7 (from eq.6b) and ABCDEF = 878008