Puzzle for May 10, 2023  ( )

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

eq.1) A + B + C + D + E + F = 22 eq.2) C = A + D eq.3) F = B + E eq.4) A + E = B + F eq.5) C + D = A + B + F eq.6) B + E + F = C + D - B

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

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Hint #1


In eq.4, replace F with B + E (from eq.3): A + E = B + B + E which becomes A + E = 2×B + E Subtract E from each side of the equation above: A + E - E = 2×B + E - E which makes A = 2×B


  

Hint #2


In eq.5, replace A with 2×B: C + D = 2×B + B + F which becomes eq.5a) C + D = 3×B + F


  

Hint #3


In eq.6, replace C + D with 3×B + F (from eq.5a): B + E + F = 3×B + F - B which becomes B + E + F = 2×B + F Subtract B and F from both sides of the above equation: B + E + F - B - F = 2×B + F - B - F which simplifies to E = B


  

Hint #4


Substitute B for E in eq.3: F = B + B which makes F = 2×B


  

Hint #5


Substitute 2×B for A in eq.2: eq.2a) C = 2×B + D


  

Hint #6


Substitute 2×B + D for C (from eq.2a), and 2×B for A and F in eq.5: 2×B + D + D = 2×B + B + 2×B which becomes 2×B + 2×D = 5×B Subtract 2×B from both sides of the equation above: 2×B + 2×D - 2×B = 5×B - 2×B which makes 2×D = 3×B Divide both sides by 2: 2×D ÷ 2 = 3×B ÷ 2 which makes D = 1½×B


  

Hint #7


Substitute 1½×B for D in eq.2a: C = 2×B + 1½×B which makes C = 3½×B


  

Solution

Substitute 2×B for A and F, 3½×B for C, 1½×B for D, and B for E in eq.1: 2×B + B + 3½×B + 1½×B + B + 2×B = 22 which simplifies to 11×B = 22 Divide both sides of the above equation by 11: 11×B ÷ 11 = 22 ÷ 11 which means B = 2 making A = F = 2×B = 2 × 2 = 4 C = 3½×B = 3½ × 2 = 7 D = 1½×B = 1½ × 2 = 3 E = B = 2 making ABDCEF = 427324