Puzzle for May 21, 2021  ( )

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

eq.1) A + B + C + D + E + F = 27 eq.2) D = C + E eq.3) A + D = C + F eq.4) C + F – B = B – C eq.5) B – A = C + E – B eq.6) A + B = average (A, D, E)

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

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Hint #1


Add B and A to both sides of eq.5: B – A + B + A = C + E – B + B + A which becomes 2×B = C + E + A which may be written as eq.5a) 2×B = A + C + E


  

Hint #2


In eq.3, replace D with C + E (from eq.2): eq.3a) A + C + E = C + F


  

Hint #3


In eq.5a, replace A + C + E with C + F (from eq.3a): eq.5b) C + F = 2×B


  

Hint #4


In eq.4, substitute 2×B for C + F (from eq.5b): 2×B – B = B – C which becomes B = B – C Subtract B from both sides of the above equation: B – B = B – C – B which makes 0 = –C which means 0 = C


  

Hint #5


Substitute 0 for C in eq.2: D = 0 + E which makes D = E


  

Hint #6


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


  

Hint #7


Substitute E for D in eq.6a: 3×A + 3×B = A + E + E which becomes 3×A + 3×B = A + 2×E Subtract A from each side of the equation above: 3×A + 3×B – A = A + 2×E – A which becomes eq.6b) 2×A + 3×B = 2×E


  

Hint #8


Substitute 0 for C in eq.5a: 2×B = A + 0 + E which becomes 2×B = A + E Subtract A from each side of the equation above: 2×B – A = A + E – A which becomes eq.5c) 2×B – A = E


  

Hint #9


Substitute (2×B – A) for E (from eq.5c) in eq.6b: 2×A + 3×B = 2×(2×B – A) which becomes 2×A + 3×B = 4×B – 2×A In the above equation, subtract 3×B from both sides, and add 2×A to both sides: 2×A + 3×B – 3×B + 2×A = 4×B – 2×A – 3×B + 2×A which makes 4×A = B


  

Hint #10


Substitute 0 for C, and (4×A) for B in eq.5b: 0 + F = 2×(4×A) which makes F = 8×A


  

Hint #11


Substitute (4×A) for B in eq.5c: 2×(4×A) – A = E which becomes 8×A – A = E which makes 7×A = E and also makes D = E = 7×A


  

Solution

Substitute 4×A for B, 0 for C, 7×A for D and E, and 8×A for F in eq.1: A + 4×A + 0 + 7×A + 7×A + 8×A = 27 which simplifies to 27×A = 27 Divide both sides of the above equation by 27: 27×A ÷ 27 = 27 ÷ 27 which means A = 1 making B = 4×A = 4 × 1 = 4 D = E = 7×A = 7 × 1 = 7 F = 8×A = 8 × 1 = 8 and ABCDEF = 140778