Puzzle for March 16, 2021  ( )

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

eq.1) A + B + C + D + E + F = 29 eq.2) E + F = A eq.3) D = C + F eq.4) A + E = C – A eq.5) C – F = A + F eq.6) B + E = D – E

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

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Hint #1


Add A to both sides of eq.4: A + E + A = C – A + A which becomes 2×A + E = C In the above equation, substitute (E + F) for A (from eq.2): 2×(E + F) + E = C which is equivalent to 2×E + 2×F + E = C which becomes eq.4a) 3×E + 2×F = C


  

Hint #2


In eq.5, replace C with 3×E + 2×F (from eq.4a), and A with E + F (from eq.2): 3×E + 2×F – F = E + F + F which becomes 3×E + F = E + 2×F Subtract E and F from both sides of the equation above: 3×E + F – E – F = E + 2×F – E – F which makes 2×E = F


  

Hint #3


In eq.2, replace F with 2×E: E + 2×E = A which makes 3×E = A


  

Hint #4


In eq.4a, replace F with (2×E): 3×E + 2×(2×E) = C which becomes 3×E + 4×E = C which makes 7×E = C


  

Hint #5


In eq.3, substitute 7×E for C, and 2×E for F: D = 7×E + 2×E which makes D = 9×E


  

Hint #6


Substitute 9×E for D in eq.6: B + E = 9×E – E which becomes B + E = 8×E Subtract E from each side of the equation above: B + E – E = 8×E – E which makes B = 7×E


  

Solution

Substitute 3×E for A, 7×E for B and C, 9×E for D, and 2×E for F in eq.1: 3×E + 7×E + 7×E + 9×E + E + 2×E = 29 which simplifies to 29×E = 29 Divide both sides of the equation above by 29: 29×E ÷ 29 = 29 ÷ 29 which means E = 1 making A = 3×E = 3 × 1 = 3 B = C = 7×E = 7 × 1 = 7 D = 9×E = 9 × 1 = 9 F = 2×E = 2 × 1 = 2 and ABCDEF = 377912