Puzzle for September 15, 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) C = A + F eq.3) B + D = E eq.4) E + F = A + D eq.5) D + E = C + F eq.6) C + D = A + B + F

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

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Hint #1


eq.6 may be written as: C + D = A + F + B In the above equation, replace A + F with C (from eq.2): C + D = C + B Subtract C from both sides: C + D – C = C + B – C which makes D = B


  

Hint #2


In eq.3, replace D with B: B + B = E which makes eq.3a) 2×B = E


  

Hint #3


In eq.4, substitute 2×B for E (from eq.3a), and B for D: 2×B + F = A + B Subtract B from both sides of the equation above: 2×B + F – B = A + B – B which becomes eq.4a) B + F = A


  

Hint #4


Substitute B for D, 2×B for E (from eq.3a), and A + F for C (from eq.2) in eq.5: B + 2×B = A + F + F which becomes eq.5a) 3×B = A + 2×F


  

Hint #5


Substitute B + F for A (from eq.4a) in eq.5a: 3×B = B + F + 2×F which becomes 3×B = B + 3×F Subtract B from both sides of the above equation: 3×B – B = B + 3×F – B which becomes 2×B = 3×F Divide both sides by 2: 2×B ÷ 2 = 3×F ÷ 2 which makes B = 1½×F and also makes D = B = 1½×F


  

Hint #6


Substitute (1½×F) for B in eq.3a: 2×(1½×F) = E which makes 3×F = E


  

Hint #7


Substitute 1½×F for B in eq.4a: 1½×F + F = A which makes 2½×F = A


  

Hint #8


Substitute 2½×F for A in eq.2: C = 2½×F + F which makes C = 3½×F


  

Solution

Substitute 2½×F for A, 1½×F for B and D, 3½×F for C, and 3×F for E in eq.1: 2½×F + 1½×F + 3½×F + 1½×F + 3×F + F = 26 which simplifies to 13×F = 26 Divide both sides of the above equation by 13: 13×F ÷ 13 = 26 ÷ 13 which means F = 2 making A = 2½×F = 2½ × 2 = 5 B = D = 1½×F = 1½ × 2 = 3 C = 3½×F = 3½ × 2 = 7 E = 3×F = 3 × 2 = 6 and ABCDEF = 537362