Puzzle for December 30, 2021  ( )

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

eq.1) A + B + C + D + E + F = 26 eq.2) B – E = E – D eq.3) E – F = D + F eq.4) A + D = B + C eq.5) C + D = A + F eq.6) D + E + F = A + B

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

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Hint #1


Subtract D from both sides of eq.4: A + D – D = B + C – D which becomes A = B + C – D In eq.5, replace A with B + C – D: C + D = B + C – D + F In the equation above, subtract C from both sides, and add D to both sides: C + D – C + D = B + C – D + F – C + D which becomes eq.5a) 2×D = B + F


  

Hint #2


Add E to both sides of eq.2: B – E + E = E – D + E which becomes eq.2a) B = 2×E – D   In eq.5a, replace B with 2×E – D (from eq.2a): 2×D = 2×E – D + F Add D to both sides of the equation above: 2×D + D = 2×E – D + F + D which becomes eq.5b) 3×D = 2×E + F


  

Hint #3


Add F to both sides of eq.3: E – F + F = D + F + F which becomes E = D + 2×F In eq.5b, substitute (D + 2×F) for E: 3×D = 2×(D + 2×F) + F which becomes 3×D = 2×D + 4×F + F which becomes 3×D = 2×D + 5×F Subtract 2×D from each side of the above equation: 3×D – 2×D = 2×D + 5×F – 2×D which makes D = 5×F


  

Hint #4


Substitute (5×F) for D in eq.5b: 3×(5×F) = 2×E + F which becomes 15×F = 2×E + F Subtract F from each side of the equation above: 15×F – F = 2×E + F – F which makes 14×F = 2×E Divide both sides by 2: 14×F ÷ 2 = 2×E ÷ 2 which makes 7×F = E


  

Hint #5


Substitute (7×F) for E, and 5×F for D in eq.2a: B = 2×(7×F) – 5×F which becomes B = 14×F – 5×F which makes B = 9×F


  

Hint #6


Substitute 5×F for D, 7×F for E, and 9×F for B in eq.6: 5×F + 7×F + F = A + 9×F which becomes 13×F = A + 9×F Subtract 9×F from both sides of the equation above: 13×F – 9×F = A + 9×F – 9×F which makes 4×F = A


  

Hint #7


Substitute 5×F for D, and 4×F for A in eq.5: C + 5×F = 4×F + F which becomes C + 5×F = 5×F Subtract 5×F from each side of the equation above: C + 5×F – 5×F = 5×F – 5×F which makes C = 0


  

Solution

Substitute 4×F for A, 9×F for B, 0 for C, 5×F for D, and 7×F for E in eq.1: 4×F + 9×F + 0 + 5×F + 7×F + F = 26 which simplifies to 26×F = 26 Divide both sides of the equation above by 26: 26×F ÷ 26 = 26 ÷ 26 which means F = 1 making A = 4×F = 4 × 1 = 4 B = 9×F = 9 × 1 = 9 D = 5×F = 5 × 1 = 5 E = 7×F = 7 × 1 = 7 and ABCDEF = 490571