Puzzle for December 16, 2023  ( )

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

eq.1) A + B + C + D + E + F = 39 eq.2) B + D = E + F eq.3) A + D = B + E - F eq.4) A + B = C + D - B eq.5) C + E - F = A + B + F eq.6) D = average (A, B, C, E)

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

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Hint #1


In eq.5, replace A + B with C + D - B (from eq.4): C + E - F = C + D - B + F In the above equation, subtract C from both sides, and add F and B to both sides: C + E - F - C + F + B = C + D - B + F - C + F + B which simplifies to E + B = D + 2×F which may be written as: eq.5a) B + E = D + 2×F


  

Hint #2


In eq.3, replace B + E with D + 2×F (from eq.5a): A + D = D + 2×F - F which becomes A + D = D + F Subtract D from each side of the equation above: A + D - D = D + F - D which makes A = F


  

Hint #3


In eq.5, substitute F for A: C + E - F = F + B + F which becomes C + E - F = 2×F + B Add F to both sides of the above equation: C + E - F + F = 2×F + B + F which becomes eq.5b) C + E = 3×F + B


  

Hint #4


eq.6 may be written as: D = (A + B + C + E) ÷ 4 Multiply both sides of the above equation by 4: 4 × D = 4 × (A + B + C + E) ÷ 4 which becomes 4×D = A + B + C + E which may be written as eq.6a) 4×D = C + E + A + B


  

Hint #5


Substitute 3×F + B for C + E (from eq.5b), and F for A in eq.6a: 4×D = 3×F + B + F + B which becomes 4×D = 4×F + 2×B Subtract 4×F from each side of the above equation: 4×D - 4×F = 4×F + 2×B - 4×F which becomes 4×D - 4×F = 2×B Divide both sides by 2: (4×D - 4×F) ÷ 2 = 2×B ÷ 2 which becomes eq.6b) 2×D - 2×F = B


  

Hint #6


Substitute 2×D - 2×F for B (from eq.6b) into eq.5a: 2×D - 2×F + E = D + 2×F In the equation above, subtract 2×D from both sides, and add 2×F to both sides: 2×D - 2×F + E - 2×D + 2×F = D + 2×F - 2×D + 2×F which becomes eq.5c) E = 4×F - D


  

Hint #7


Substitute 2×D - 2×F for B (from eq.6b) and 4×F - D for E (from eq.5c) in eq.2: 2×D - 2×F + D = 4×F - D + F which becomes 3×D - 2×F = 5×F - D Add 2×F and D to both sides of the equation above: 3×D - 2×F + 2×F + D = 5×F - D + 2×F + D which becomes 4×D = 7×F Divide both sides by 4: 4×D ÷ 4 = 7×F ÷ 4 which makes D = 1¾×F


  

Hint #8


Substitute 1¾×F for D in eq.5c: E = 4×F - 1¾×F which makes E = 2¼×F


  

Hint #9


Substitute (1¾×F) for D in eq.6b: 2×(1¾×F) - 2×F = B which becomes 3½×F - 2×F = B which makes 1½×F = B


  

Hint #10


Substitute 2¼×F for E, and 1½×F for B in eq.5b: C + 2¼×F = 3×F + 1½×F which becomes C + 2¼×F = 4½×F Subtract 2¼×F from each side of the equation above: C + 2¼×F - 2¼×F = 4½×F - 2¼×F which makes C = 2¼×F


  

Solution

Substitute F for A, 1½×F for B, 2¼×F for C and E, and 1¾×F for D in eq.1: F + 1½×F + 2¼×F + 1¾×F + 2¼×F + F = 39 which simplifies to 9¾×F = 39 Divide both sides of the above equation by 9¾: 9¾×F ÷ 9¾ = 39 ÷ 9¾ which means F = 4 making A = F = 4 B = 1½×F = 1½ × 4 = 6 C = E = 2¼×F = 2¼ × 4 = 9 D = 1¾×F = 1¾ × 4 = 7 and ABCDEF = 469794