Puzzle for November 24, 2022  ( )

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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 + E = A + B eq.3) B + E = A - C + D eq.4) E - D = A - B + C eq.5) B - D = A + D - F eq.6)* A = BC ÷ D

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


Add D and B to both sides of eq.4: E - D + D + B = A - B + C + D + B which becomes eq.4a) E + B = A + C + D


  

Hint #2


Subtract the left and right sides of eq.3 from the left and right sides of eq.4a, respectively: E + B - (B + E) = A + C + D - (A - C + D) which becomes E + B - B - E = A + C + D - A + C - D which simplifies to 0 = 2×C which means 0 = C


  

Hint #3


In eq.2, replace C with 0: 0 + E = A + B which becomes eq.2a) E = A + B


  

Hint #4


In eq.4, replace E with A + B (from eq.2a), and C with 0: A + B - D = A - B + 0 which becomes A + B - D = A - B In the above equation, subtract A from both sides, and add D and B to both sides: A + B - D - A + D + B = A - B - A + D + B which simplifies to 2×B = D


  

Hint #5


eq.6 may be written as: A = (10×B + C) ÷ D In the above equation, replace C with 0, and D with 2×B: A = (10×B + 0) ÷ 2×B which becomes A = 10×B ÷ 2×B which makes A = 5


  

Hint #6


Substitute 5 for A in eq.2a: eq.2b) E = 5 + B


  

Hint #7


Substitute 2×B for D, and 5 for A in eq.5: B - 2×B = 5 + 2×B - F which becomes -B = 5 + 2×B - F Add B and F to both sides of the equation above: -B + B + F = 5 + 2×B - F + B + F which becomes eq.5a) F = 5 + 3×B


  

Hint #8


Substitute 5 for A, 0 for C, 2×B for D, 5 + B for E (from eq.2b), and 5 + 3×B for F (from eq.5a) in eq.1: 5 + B + 0 + 2×B + 5 + B + 5 + 3×B = 22 which simplifies to 15 + 7×B = 22 Subtract 15 from each side of the above equation: 15 + 7×B - 15 = 22 - 15 which makes 7×B = 7 Divide both sides by 7: 7×B ÷ 7 = 7 ÷ 7 which makes B = 1


  

Solution

Since B = 1, then: D = 2×B = 2 × 1 = 2 E = 5 + B = 5 + 1 = 6 (from eq.2b) F = 5 + 3×B = 5 + 3×1 = 5 + 3 = 8 (from eq.5a) and ABCDEF = 510268