Puzzle for November 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 = 28 eq.2) D = A + B eq.3) C = D + F eq.4) B + E = C – B eq.5) A + F = B + D eq.6) D + E – F = B + F

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

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Hint #1


In eq.5, replace D with A + B (from eq.2): A + F = B + A + B which becomes A + F = 2×B + A Subtract A from each side of the equation above: A + F – A = 2×B + A – A which makes F = 2×B


  

Hint #2


Add B to both sides of eq.4: B + E + B = C – B + B which becomes 2×B + E = C In the equation above, replace 2×B with F: eq.4a) F + E = C


  

Hint #3


In eq.3, substitute F + E for C (from eq.4a): F + E = D + F Subtract F from each side of the above equation: F + E – F = D + F – F which makes E = D


  

Hint #4


Substitute D for E, and 2×B for F in eq.6: D + D – 2×B = B + 2×B which becomes 2×D – 2×B = 3×B Add 2×B to both sides of the above equation: 2×D – 2×B + 2×B = 3×B + 2×B which makes 2×D = 5×B Divide both sides by 2: 2×D ÷ 2 = 5×B ÷ 2 which makes D = 2½×B and also makes E = D = 2½×B


  

Hint #5


Substitute 2½×B for D in eq.2: 2½×B = A + B Subtract B from each side of the equation above: 2½×B – B = A + B – B which makes 1½×B = A


  

Hint #6


Substitute 2×B for F, and 2½×B for E in eq.4a: 2×B + 2½×B = C which makes 4½×B = C


  

Solution

Substitute 1½×B for A, 4½×B for C, 2½×B for D and E, and 2×B for F in eq.1: 1½×B + B + 4½×B + 2½×B + 2½×B + 2×B = 28 which simplifies to 14×B = 28 Divide both sides of the above equation by 14: 14×B ÷ 14 = 28 ÷ 14 which means B = 2 making A = 1½×B = 1½ × 2 = 3 C = 4½×B = 4½ × 2 = 9 D = E = 2½×B = 2½ × 2 = 5 F = 2×B = 2 × 2 = 4 and ABCDEF = 329554