Puzzle for March 24, 2020  ( )

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

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

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Hint #1


In eq.2, replace A + B with D + E (from eq.5): C + E = D + E – C In the equation above, subtract E from each side, and add C to both sides: C + E – E + C = D + E – C – E + C which makes 2×C = D


  

Hint #2


In eq.5, replace B with A + D (from eq.3): D + E = A + A + D which becomes D + E = 2×A + D Subtract D from each side of the above equation: D + E – D = 2×A + D – D which makes eq.3a) E = 2×A


  

Hint #3


Add C to both sides of eq.6: B – C + C = C + D + C which becomes B = 2×C + D In eq.3, substitute 2×C + D for B: 2×C + D = A + D Subtract D from both sides of the equation above: 2×C + D – D = A + D – D which makes 2×C = A


  

Hint #4


Substitute (2×C) for A in eq.3a: E = 2×(2×C) which makes E = 4×C


  

Hint #5


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


  

Hint #6


Substitute 2×C for D, and 2×C for A in eq.4: C – 2×C = 2×C – F which becomes –C = 2×C – F Add C and F to both sides of the above equation: –C + C + F = 2×C – F + C + F which makes F = 3×C


  

Solution

Substitute 2×C for A and D, 4×C for B and E, and 3×C for F in eq.1: 2×C + 4×C + C + 2×C + 4×C + 3×C = 32 which simplifies to 16×C = 32 Divide both sides of the equation above by 16: 16×C ÷ 16 = 32 ÷ 16 which means C = 2 making A = D = 2×C = 2 × 2 = 4 B = E = 4×C = 4 × 2 = 8 F = 3×C = 3 × 2 = 6 and ABCDEF = 482486