Puzzle for August 31, 2023  ( )

Scratchpad

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

eq.1) A + B + C + D + E + F = 27 eq.2) C - D = A - C eq.3) B - E = E - F eq.4) E - D = C - E eq.5) F - D = D - C eq.6) A - D = E

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

Scratchpad

 

Help Area

Hint #1


Add D to both sides of eq.6: A - D + D = E + D which becomes eq.6a) A = E + D


  

Hint #2


In eq.2, replace A with E + D (from eq.6a): C - D = E + D - C In the above equation, subtract D from both sides, and add C to both sides: C - D - D + C = E + D - C - D + C which becomes eq.2a) 2×C - 2×D = E


  

Hint #3


In eq.4, substitute (2×C - 2×D) for E (from eq.2a): (2×C - 2×D) - D = C - (2×C - 2×D) which becomes 2×C - 3×D = C - 2×C + 2×D which becomes 2×C - 3×D = -C + 2×D Add 3×D and C to both sides of the above equation: 2×C - 3×D + 3×D + C = -C + 2×D + 3×D + C which makes eq.4a) 3×C = 5×D


  

Hint #4


Add D to both sides of eq.5: F - D + D = D - C + D which becomes F = 2×D - C Multiply both sides of the above equation by 3: 3×F = 3×(2×D - C) which becomes eq.5a) 3×F = 6×D - 3×C


  

Hint #5


In eq.5a, replace 3×C with 5×D: 3×F = 6×D - 5×D which makes 3×F = D


  

Hint #6


In eq.4a, substitute (3×F) for D: 3×C = 5×(3×F) which makes 3×C = 15×F Divide both sides of the above equation by 3: 3×C ÷ 3 = 15×F ÷ 3 which makes C = 5×F


  

Hint #7


Substitute 5×F for C, and 3×F for D in eq.2a: 2×(5×F) - 2×(3×F) = E which becomes 10×F - 6×F = E which makes 4×F = E


  

Hint #8


Substitute 4×F for E in eq.3: B - 4×F = 4×F - F which becomes B - 4×F = 3×F Add 4×F to both sides of the equation above: B - 4×F + 4×F = 3×F + 4×F which makes B = 7×F


  

Hint #9


Substitute 4×F for E, and 3×F for D in eq.6a: A = 4×F + 3×F which makes A = 7×F


  

Solution

Substitute 7×F for A and B, 5×F for C, 3×F for D, and 4×F for E in eq.1: 7×F + 7×F + 5×F + 3×F + 4×F + F = 27 which simplifies to 27×F = 27 Divide both sides of the above equation by 27: 27×F ÷ 27 = 27 ÷ 27 which means F = 1 making A = B = 7×F = 7 × 1 = 7 C = 5×F = 5 × 1 = 5 D = 3×F = 3 × 1 = 3 E = 4×F = 4 × 1 = 4 and ABCDEF = 775341