Puzzle for August 9, 2022  ( )

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

eq.1) A + B + C + D + E + F = 24 eq.2) D = A + B eq.3) E = B + C eq.4) A = C + F eq.5) A + E = B + D eq.6) B + E – A = A + 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 + E = B + A + B which becomes A + E = 2×B + A Subtract A from each side of the equation above: A + E – A = 2×B + A – A which makes eq.5a) E = 2×B


  

Hint #2


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


  

Hint #3


In eq.4, replace C with B: eq.4a) A = B + F


  

Hint #4


In eq.6, substitute 2×B for E, and (B + F) for A (from eq.4a): B + 2×B – (B + F) = (B + F) + F which becomes 3×B – B – F = B + 2×F which becomes 2×B – F = B + 2×F In the above equation, add F to both sides, and subtract B from both sides: 2×B – F + F – B = B + 2×F + F – B which makes B = 3×F and also makes B = C = 3×F


  

Hint #5


Substitute (3×F) for B in eq.5a: E = 2×(3×F) which makes E = 6×F


  

Hint #6


Substitute 3×F for B in eq.4a: A = 3×F + F which makes A = 4×F


  

Hint #7


Substitute 4×F for A, and 3×F for B in eq.2: D = 4×F + 3×F which makes D = 7×F


  

Solution

Substitute 4×F for A, 3×F for B and C, 7×F for D, and 6×F for E in eq.1: 4×F + 3×F + 3×F + 7×F + 6×F + F = 24 which simplifies to 24×F = 24 Divide both sides of the above equation by 24: 24×F ÷ 24 = 24 ÷ 24 which means F = 1 making A = 4×F = 4 × 1 = 4 B = C = 3×F = 3 × 1 = 3 D = 7×F = 7 × 1 = 7 E = 6×F = 6 × 1 = 6 and ABCDEF = 433761