Puzzle for August 13, 2021  ( )

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

eq.1) A + B + C + D + E + F = 24 eq.2) B = average (D, E) eq.3) C = average (D, E, F) eq.4) F = average (B, C, D, E) eq.5) D + (A ÷ B) = E + (C ÷ F) eq.6) A × B = (B × C) + (C × D) + (E × F)

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

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Hint #1


eq.2 may be written as: B = (D + E) ÷ 2 Multiply both sides of the above equation by 2: 2 × B = 2 × ((D + E) ÷ 2) which becomes eq.2a) 2×B = D + E


  

Hint #2


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


  

Hint #3


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


  

Hint #4


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


  

Hint #5


In eq.4a, replace D + E with 2×B (from eq.2a): 4×F = B + C + 2×B which becomes 4×F = 3×B + C Subtract 3×B from each side of the equation above: 4×F – 3×B = 3×B + C – 3×B which becomes eq.4b) 4×F – 3×B = C


  

Hint #6


In eq.3b, substitute (4×F – 3×B) for C (from eq.4b): 3×(4×F – 3×B) = 2×B + F which is equivalent to 12×F – 9×B = 2×B + F In the equation above, add 9×B to both sides, and subtract F from both sides: 12×F – 9×B + 9×B – F = 2×B + F + 9×B – F which becomes 11×F = 11×B Divide both sides by 11: 11×F ÷ 11 = 11×B ÷ 11 which makes F = B


  

Hint #7


Substitute B for F in eq.4b: 4×B – 3×B = C which makes B = C


  

Hint #8


Substitute B for C and F in eq.6: A × B = (B × B) + (B × D) + (E × B) which may be written as A × B = B × (B + D + E) Since B ≠ 0 (from eq.5), divide both sides of the above equation by B: (A × B) ÷ B = (B × (B + D + E)) ÷ B which becomes eq.6a) A = B + D + E


  

Hint #9


Substitute 2×B for D + E (from eq.2a) into eq.6a: A = B + 2×B which makes A = 3×B


  

Hint #10


Substitute 3×B for A, B for C and F, and 2×B for D + E (from eq.2a) in eq.1: 3×B + B + B + 2×B + B = 24 which makes 8×B = 24 Divide both sides of the above equation by 8: 8×B ÷ 8 = 24 ÷ 8 which means B = 3 and makes C = F = B = 3 A = 3×B = 3 × 3 = 9


  

Hint #11


Substitute 9 for A, and 3 for B and C and F in eq.5: D + (9 ÷ 3) = E + (3 ÷ 3) which becomes D + 3 = E + 1 Subtract 1 from both sides of the equation above: D + 3 – 1 = E + 1 – 1 which makes eq.5a) D + 2 = E


  

Hint #12


Substitute 3 for B, and D + 2 for E (from eq.5a) in eq.2a: 2×3 = D + D + 2 which becomes 6 = 2×D + 2 Subtract 2 from both sides of the above equation: 6 – 2 = 2×D + 2 – 2 which makes 4 = 2×D Divide both sides by 2: 4 ÷ 2 = 2×D ÷ 2 which makes 2 = D


  

Solution

Substitute 2 for D in eq.5a: 2 + 2 = E which makes 4 = E and makes ABCDEF = 933243