Puzzle for March 4, 2021  ( )

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

eq.1) A + B + C + D + E + F = 14 eq.2) C + E = A eq.3) D + F = A + C eq.4) A + C + F = B + D eq.5) B + C = D + E eq.6) C = average (A, D, 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 A + C with D + F (from eq.3): D + F + F = B + D which becomes D + 2×F = B + D Subtract D from each side of the above equation: D + 2×F – D = B + D – D which makes 2×F = B


  

Hint #2


eq.6 may be written as: C = (A + D + F) ÷ 3 Multiply both sides of the equation above by 3: 3 × C = 3 × (A + D + F) ÷ 3 which becomes 3×C = A + D + F Subtract A from both sides: 3×C – A = A + D + F – A which becomes eq.6a) 3×C – A = D + F


  

Hint #3


In eq.3, replace D + F with 3×C – A (from eq.6a): 3×C – A = A + C In the equation above, add A to both sides, and subtract C from both sides: 3×C – A + A – C = A + C + A – C which becomes 2×C = 2×A Divide both sides by 2: 2×C ÷ 2 = 2×A ÷ 2 which makes C = A


  

Hint #4


In eq.2, substitute A for C: A + E = A Subtract A from both sides of the above equation: A + E – A = A – A which makes E = 0


  

Hint #5


Substitute 0 for E in eq.5: B + C = D + 0 which makes eq.5a) B + C = D


  

Hint #6


Substitute B + C for D (from eq.5a) in eq.4: A + C + F = B + B + C which becomes A + C + F = 2×B + C Subtract C from each side of the above equation: A + C + F – C = 2×B + C – C which becomes eq.4a) A + F = 2×B


  

Hint #7


Substitute (2×F) for B in eq.4a: A + F = 2×(2×F) which becomes A + F = 4×F Subtract F from each side of the equation above: A + F – F = 4×F – F which makes A = 3×F and also makes C = A = 3×F


  

Hint #8


Substitute 2×F for B, and 3×F for C in eq.5a: 2×F + 3×F = D which makes 5×F = D


  

Solution

Substitute 3×F for A and C, 2×F for B, 5×F for D, and 0 for E in eq.1: 3×F + 2×F + 3×F + 5×F + 0 + F = 14 which simplifies to 14×F = 14 Divide both sides of the equation above by 14: 14×F ÷ 14 = 14 ÷ 14 which means F = 1 making A = C = 3×F = 3 × 1 = 3 B = 2×F = 2 × 1 = 2 D = 5×F = 5 × 1 = 5 and ABCDEF = 323501