Puzzle for July 7, 2021  ( )

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

eq.1) A + B + C + D + E + F = 21 eq.2) D = B + F eq.3) B + C = A + D eq.4) C – F = A + F eq.5) B – E + F = A + C + E eq.6) E = average (C, D, F)

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

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Hint #1


Add F to both sides of eq.4: C – F + F = A + F + F which becomes eq.4a) C = A + 2×F   In eq.3, replace C with A + 2×F (from eq.4a), and D with B + F (from eq.2): B + A + 2×F = A + B + F Subtract A, B, and F from both sides of the above equation: B + A + 2×F – A – B – F = A + B + F – A – B – F which simplifies to F = 0


  

Hint #2


In eq.2, replace F with 0: D = B + 0 which makes D = B


  

Hint #3


In eq.4a, replace F with 0: C = A + 2×0 which makes C = A


  

Hint #4


In eq.5, substitute 0 for F, and A for C: B – E + 0 = A + A + E which becomes B – E = 2×A + E Add E to both sides of the above equation: B – E + E = 2×A + E + E which becomes eq.5a) B = 2×A + 2×E


  

Hint #5


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


  

Hint #6


In eq.6a, substitute A for C, B for D, and 0 for F: 3×E = A + B + 0 which becomes eq.6b) 3×E = A + B


  

Hint #7


Substitute 2×A + 2×E for B (from eq.5a) in eq.6a: 3×E = A + 2×A + 2×E which becomes 3×E = 3×A + 2×E Subtract 2×E from each side of the above equation: 3×E – 2×E = 3×A + 2×E – 2×E which makes E = 3×A


  

Hint #8


Substitute (3×A) for E in eq.5a: B = 2×A + 2×(3×A) which becomes B = 2×A + 6×A which makes B = 8×A and also makes D = B = 8×A


  

Solution

Substitute 8×A for B and D, A for C, 3×A for E, and 0 for F in eq.1: A + 8×A + A + 8×A + 3×A + 0 = 21 which simplifies to 21×A = 21 Divide both sides of the above equation by 21: 21×A ÷ 21 = 21 ÷ 21 which means A = 1 making B = D = 8×A = 8 × 1 = 8 C = A = 1 E = 3×A = 3 × 1 = 3 and ABCDEF = 181830