Puzzle for September 25, 2021  ( )

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

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

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

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Hint #1


In eq.2, replace A + E with F – E (from eq.1): B – F = F – E + F which becomes B – F = 2×F – E Add F and E to both sides of the above equation: B – F + F + E = 2×F – E + F + E which becomes eq.2a) B + E = 3×F


  

Hint #2


eq.5 may be written as: F = (B + D + E) ÷ 3 Multiply both sides of the equation above by 3: F × 3 = ((B + D + E) ÷ 3) × 3 which becomes eq.5a) 3×F = B + D + E


  

Hint #3


In eq.5a, replace 3×F with B + E (from eq.2a): B + E = B + D + E Subtract B and E from both sides of the equation above: B + E – B – E = B + D + E – B – E which makes 0 = D


  

Hint #4


In eq.3, substitute 0 for D: 0 + E – A = A – E which becomes E – A = A – E Add A and E to both sides of the above equation: E – A + A + E = A – E + A + E which becomes 2×E = 2×A Divide both sides by 2: 2×E ÷ 2 = 2×A ÷ 2 which makes A = E


  

Hint #5


Substitute E for A in eq.1: F – E = E + E which becomes F – E = 2×E Add E to both sides of the equation above: F – E + E = 2×E + E which makes F = 3×E


  

Hint #6


Substitute 3×E for F, and 0 for D in eq.5a: 3×(3×E) = B + 0 + E which becomes 9×E = B + E Subtract E from each side of the above equation: 9×E – E = B + E – E which makes 8×E = B


  

Hint #7


Substitute 8×E for B, 3×E for F, and E for A in eq.4: 8×E + E = 3×E ^ (E + E) which becomes 9×E = 3×E ^ (E + E) which may be written as 3 × 3×E = (3×E ^ E) × (3×E ^ E) Since E ≠ 0 (from eq.6), divide both sides of the above equation by 3×E: 3 × 3×E ÷ 3×E = (3×E ^ E) × (3×E ^ E) ÷ 3×E which becomes 3 = (1 ^ E) × (3×E ^ E) which becomes eq.4a) 3 = 3×E ^ E


  

Hint #8


To make eq.4a true, check several possible values for E:   If E = 0, then 3×0 ^ 0 = 0 ^ 0 = unknown If E = 1, then 3×1 ^ 1 = 3 ^ 1 = 3 If E = 2, then 3×2 ^ 2 = 6 ^ 2 = 36 If E > 2, then E > 36   Since E is a one-digit integer, the above equations make: E = 1 and also make A = E = 1 B = 8×E = 8 × 1 = 8 F = 3×E = 3 × 1 = 3


  

Solution

Substitute 3 for F, 0 for D, and 1 for A and E in eq.6: 3! – (0 ÷ 1) = C + 0 – 1 – 1 which becomes 6 – 0 = C – 2 which means 6 = C – 2 Add 2 to both sides of the above equation: 6 + 2 = C – 2 + 2 which makes 8 = C and makes ABCDEF = 188013