Puzzle for March 10, 2022  ( )

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

eq.1) A + B + C + D + E + F = 32 eq.2) E = A + F eq.3) D = B + C eq.4) A + B = D + E – B – F eq.5) E + F – A = A + C – F eq.6) C + D – A – F = A – C + F

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

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Hint #1


In eq.4, replace D with B + C (from eq.3), and E with A + F (from eq.2): A + B = B + C + A + F – B – F which becomes A + B = C + A Subtract A from each side of the above equation: A + B – A = C + A – A which makes B = C


  

Hint #2


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


  

Hint #3


In eq.6, substitute B for C, and 2×B for D: B + 2×B – A – F = A – B + F which becomes 3×B – A – F = A – B + F Add A, F, and B to both sides of the above equation: 3×B – A – F + A + F + B = A – B + F + A + F + B which becomes 4×B = 2×A + 2×F which may be written as eq.6a) 4×B = 2×(A + F)


  

Hint #4


In eq.6a, replace (A + F) with E (from eq.2): 4×B = 2×E Divide both sides of the above equation by 2: 4×B ÷ 2 = 2×E ÷ 2 which makes 2×B = E


  

Hint #5


Substitute 2×B for E, and B for C in eq.5: 2×B + F – A = A + B – F In the equation above, subtract F and B from both sides, and add A to both sides: 2×B + F – A – F – B + A = A + B – F – F – B + A which becomes eq.5a) B = 2×A – 2×F


  

Hint #6


Substitute (2×A – 2×F) for B (from eq.5a) in eq.6a: 4×(2×A – 2×F) = 2×(A + F) which becomes 8×A – 8×F = 2×A + 2×F In the above equation, add 8×F to each side, and subtract 2×A from both sides: 8×A – 8×F + 8×F – 2×A = 2×A + 2×F + 8×F – 2×A which simplifies to eq.6b) 6×A = 10×F


  

Hint #7


Multiply both sides of eq.5a by 3: 3×B = 3×(2×A – 2×F) which becomes 3×B = 6×A – 6×F Substitute 10×F for 6×A (from eq.6b) into the above equation: 3×B = 10×F – 6×F which makes 3×B = 4×F Divide both sides by 4: 3×B ÷ 4 = 4×F ÷ 4 which makes ¾×B = F


  

Hint #8


Substitute (¾×B) for F in eq.5a: B = 2×A – 2×(¾×B) which becomes B = 2×A – 1½×B Add 1½×B to both sides of the equation above: B + 1½×B = 2×A – 1½×B + 1½×B which makes 2½×B = 2×A Divide both sides by 2: (2½×B) ÷ 2 = 2×A ÷ 2 which makes 1¼×B = A


  

Solution

Substitute 1¼×B for A, B for C, 2×B for D and E, and ¾×B for F in eq.1: 1¼×B + B + B + 2×B + 2×B + ¾×B = 32 which simplifies to 8×B = 32 Divide both sides of the above equation by 8: 8×B ÷ 8 = 32 ÷ 8 which means B = 4 making A = 1¼×B = 1¼ × 4 = 5 C = B = 4 D = E = 2×B = 2 × 4 = 8 F = ¾×B = ¾ × 4 = 3 and ABCDEF = 544883