Puzzle for December 28, 2023  ( )

Scratchpad

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

eq.1) A + B + C + D + E + F = 27 eq.2) F = B + E eq.3) D + E = C eq.4) D + F = B + C - E eq.5) C + D - A + E = A + B + F - E eq.6) B + C + F = A + D + (B ÷ F)

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

Scratchpad

 

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Hint #1


Add the left and right sides of eq.2 to the left and right sides of eq.3, respectively: D + E + F = C + B + E Subtract E from each side of the above equation: D + E + F - E = C + B + E - E which becomes D + F = C + B which is the same as eq.3a) D + F = B + C


  

Hint #2


In eq.4, replace B + C with D + F (from eq.3a): D + F = D + F - E Subtract D and F from each side of the equation above: D + F - D - F = D + F - E - D - F which makes 0 = -E which means 0 = E


  

Hint #3


In eq.2, substitute 0 for E: F = B + 0 which makes F = B


  

Hint #4


In eq.3, substitute 0 for E: D + 0 = C which makes D = C


  

Hint #5


Substitute C for D, 0 for E, and B for F in eq.5: C + C - A + 0 = A + B + B - 0 which becomes 2×C - A = A + 2×B Add A to both sides of the above equation: 2×C - A + A = A + 2×B + A which becomes 2×C = 2×A + 2×B Divide both sides by 2: 2×C ÷ 2 = (2×A + 2×B) ÷ 2 which becomes eq.5a) C = A + B


  

Hint #6


Substitute B for F, and C for D in eq.6: B + C + B = A + C + (B ÷ B) which becomes 2×B + C = A + C + 1 Subtract C and 1 from both sides of the equation above: 2×B + C - C - 1 = A + C + 1 - C - 1 which makes eq.6a) 2×B - 1 = A


  

Hint #7


Substitute 2×B - 1 for A (from eq.6a) into eq.5a: C = 2×B - 1 + B which makes C = 3×B - 1 and also makes eq.5b) D = C = 3×B - 1


  

Hint #8


Substitute 2×B - 1 for A (from eq.6a), 3×B - 1 for C and D (from eq.5b), 0 for E, and B for F in eq.1: 2×B - 1 + B + 3×B - 1 + 3×B - 1 + 0 + B = 27 which simplifies to 10×B - 3 = 27 Add 3 to both sides of the above equation: 10×B - 3 + 3 = 27 + 3 which makes 10×B = 30 Divide both sides by 10: 10×B ÷ 10 = 30 ÷ 10 which means B = 3


  

Solution

Since B = 3, then: A = 2×B - 1 = 2×3 - 1 = 6 - 1 = 5 (from eq.6a) C = D = 3×B - 1 = 3×3 - 1 = 9 - 1 = 8 (from eq.5b) F = B = 3 and ABCDEF = 538803