Puzzle for July 18, 2021  ( )

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

eq.1) A + B + C + D + E + F = 17 eq.2) B + C = A + D eq.3) E = B + D + F eq.4) C + D = A + B + F eq.5) D = average (A, C, E, F) eq.6)* BC – E = A + B + C + D + E

A, B, C, D, E, and F each represent a one-digit non-negative integer.
*  BC is a 2-digit number (not B×C).

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Hint #1


eq.6 may be written as: eq.6a) 10×B + C – E = A + B + C + D + E   In eq.6a above, subtract B and C from both sides, and add E to both sides: 10×B + C – E – B – C + E = A + B + C + D + E – B – C + E which simplifies to eq.6b) 9×B = A + D + 2×E


  

Hint #2


In eq.6b, replace A + D with B + C (from eq.2): 9×B = B + C + 2×E Subtract B from each side of the equation above: 9×B – B = B + C + 2×E – B which becomes eq.6c) 8×B = C + 2×E


  

Hint #3


eq.5 may be written as: D = (A + C + E + F) ÷ 4 Multiply both sides of the above equation by 4: 4 × D = 4 x (A + C + E + F) ÷ 4 which becomes eq.5a) 4×D = A + C + E + F


  

Hint #4


eq.1 may be written as: A + C + E + F + B + D = 17 In the above equation, replace A + C + E + F with 4×D (from eq.5a): 4×D + B + D = 17 which becomes eq.1a) 5×D + B = 17


  

Hint #5


In eq.1, replace A + B + C + D + E with 10×B + C – E (from eq.6a): eq.1b) 10×B + C – E + F = 17


  

Hint #6


In eq.1a, substitute 10×B + C – E + F for 17 (from eq.1b): 5×D + B = 10×B + C – E + F Subtract B from both sides of the equation above: 5×D + B – B = 10×B + C – E + F – B which becomes 5×D = 9×B + C – E + F which may be written as eq.1c) 5×D = B + 8×B + C – E + F


  

Hint #7


Substitute C + 2×E for 8×B (from eq.6c) in eq.1c: 5×D = B + C + 2×E + C – E + F which becomes 5×D = B + 2×C + E + F which may be written as eq.1d) 5×D = B + F + 2×C + E


  

Hint #8


Subtract D from each side of eq.3: E – D = B + D + F – D which becomes E – D = B + F In eq.1d, substitute E – D for B + F: 5×D = E – D + 2×C + E which becomes 5×D = 2×E – D + 2×C Add D to both sides of the above equation: 5×D + D = 2×E – D + 2×C + D which becomes 6×D = 2×E + 2×C Divide both sides by 2: 6×D ÷ 2 = (2×E + 2×C) ÷ 2 which becomes 3×D = E + C which is the same as eq.3a) 3×D = C + E


  

Hint #9


Substitute 3×D for C + E (from eq.3a) in eq.5a: 4×D = A + 3×D + F Subtract 3×D from each side of the above equation: 4×D – 3×D = A + 3×D + F – 3×D which becomes eq.5b) D = A + F


  

Hint #10


Substitute A + F for D (from eq.5b) in eq.4: C + A + F = A + B + F Subtract A and F from both sides of the equation above: C + A + F – A – F = A + B + F – A – F which simplifies to C = B


  

Hint #11


Substitute B for C in eq.6c: 8×B = B + 2×E Subtract B from both sides of the above equation: 8×B – B = B + 2×E – B which makes 7×B = 2×E Divide both sides by 2: 7×B ÷ 2 = 2×E ÷ 2 which makes 3½×B = E


  

Hint #12


Substitute B for C, and 3½×B for E in eq.3a: 3×D = B + 3½×B which makes 3×D = 4½×B Divide both sides of the above equation by 3: 3×D ÷ 3 = 4½×B ÷ 3 which makes D = 1½×B


  

Hint #13


Substitute B for C, and 1½×B for D in eq.2: B + B = A + 1½×B which becomes 2×B = A + 1½×B Subtract 1½×B from both sides of the equation above: 2×B – 1½×B = A + 1½×B – 1½×B which makes ½×B = A


  

Hint #14


Substitute 1½×B for D, and ½×B for A in eq.5b: 1½×B = ½×B + F Subtract ½×B from each side of the equation above: 1½×B – ½×B = ½×B + F – ½×B which makes B = F


  

Solution

Substitute (1½×B) for D in eq.1a: 5×(1½×B) + B = 17 which becomes 7½×B + B = 17 which makes 8½×B = 17 Divide both sides of the equation above by 8½: 8½×B ÷ 8½ = 17 ÷ 8½ which means B = 2 making A = ½×B = ½ × 2 = 1 C = F = B = 2 D = 1½×B = 1½ × 2 = 3 E = 3½×B = 3½ × 2 = 7 and ABCDEF = 122372