Puzzle for July 7, 2023  ( )

Scratchpad

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

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

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

Scratchpad

 

Help Area

Hint #1


Subtract A and E from each side of eq.2: A + C - A - E = D + E - A - E which becomes eq.2a) C - E = D - A   Subtract B and E from each side of eq.3: B + C - B - E = A + E - B - E which becomes eq.3a) C - E = A - B


  

Hint #2


In eq.3a, replace C - E with D - A (from eq.2a): D - A = A - B Add A and B to both sides of the above equation: D - A + A + B = A - B + A + B which becomes D + B = 2×A which may be written as eq.3b) B + D = 2×A


  

Hint #3


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


  

Hint #4


In eq.4a, replace B + D with 2×A (from eq.3b): 2×A + C = 2×A Subtract 2×A from each side of the equation above: 2×A + C - 2×A = 2×A - 2×A which makes C = 0


  

Hint #5


In eq.5, substitute 0 for C: E ^ F = D ^ 0 which makes eq.5a) E ^ F = 1 (implies D ≠ 0)


  

Hint #6


eq.5a implies that either: E = 1 or: E > 1 and F = 0   Since F ≠ 0 (from eq.6), then: E = 1


  

Hint #7


Substitute 0 for C, and 1 for E in eq.2: A + 0 = D + 1 which makes eq.2b) A = D + 1


  

Hint #8


Substitute 0 for C, 1 for E, and D + 1 for A (from eq.2b) in eq.3a: 0 - 1 = D + 1 - B which becomes -1 = D + 1 - B Add 1 and B to both sides of the above equation: -1 + 1 + B = D + 1 - B + 1 + B which makes eq.3c) B = D + 2


  

Hint #9


Substitute D + 2 for B (from eq.3c), and 1 for E in eq.1: D + F = D + 2 + 1 which makes D + F = D + 3 Subtract D from each side of the equation above: D + F - D = D + 3 - D which makes F = 3


  

Hint #10


Substitute D + 2 for B (from eq.3c), 1 for E, 3 for F, D + 1 for A (from eq.2b), and 0 for C in eq.6: D + 2 - 1 - 3 = (D + 1 + 0 + D) ÷ 3 which becomes D - 2 = (2×D + 1) ÷ 3 Multiply both sides of the above equation by 3: 3 × (D - 2) = 3 × (2×D + 1) ÷ 3 which becomes 3×D - 6 = 2×D + 1 Add 6 to both sides, and subtract 2×D from both sides: 3×D - 6 + 6 - 2×D = 2×D + 1 + 6 - 2×D which makes D = 7


  

Solution

Since D = 7, then: B = D + 2 = 7 + 2 = 9 (from eq.3c) A = D + 1 = 7 + 1 = 8 (from eq.2b) and ABCDEF = 890713