Puzzle for June 25, 2023 ( )
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Find the 6-digit number ABCDEF by solving the following equations:
A, B, C, D, E, and F each represent a one-digit non-negative integer.
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Hint #1
eq.5 may be written as: D - (E + F) = F - A - B In the above equation, replace E + F with B + C (from eq.2): D - (B + C) = F - A - B which becomes D - B - C = F - A - B Add B, C, and A to both sides of the above equation: D - B - C + B + C + A = F - A - B + B + C + A which simplifies to eq.5a) D + A = F + C
Hint #2
Add C to both sides of eq.4: B + C + C = A - C + D + C which becomes B + 2×C = A + D which may be written as eq.4a) B + 2×C = D + A
Hint #3
In eq.5a, replace D + A with B + 2×C (from eq.4a): B + 2×C = F + C Subtract C from each side of the equation above: B + 2×C - C = F + C - C which becomes eq.5b) B + C = F
Hint #4
In eq.2, substitute F for B + C (from eq.5b): F = E + F Subtract F from each side of the above equation: F - F = E + F - F which makes 0 = E
Hint #5
Subtract D and C from each side of eq.5a: D + A - D - C = F + C - D - C which becomes eq.5c) A - C = F - D eq.6 may be written as: eq.6a) F - A - B + E = A - C + B - E
Hint #6
In eq.6a, substitute 0 for E, and F - D for A - C (from eq.5c): F - A - B + 0 = F - D + B - 0 which becomes F - A - B = F - D + B In the equation above, subtract F from both sides, and add A and B and D to both sides: F - A - B - F + A + B + D = F - D + B - F + A + B + D which simplifies to eq.6b) D = 2×B + A
Hint #7
Substitute 2×B + A for D (from eq.6b), and 0 for E in eq.5: 2×B + A - 0 - F = F - A - B which becomes 2×B + A - F = F - A - B Add F, A, and B to both sides of the above equation: 2×B + A - F + F + A + B = F - A - B + F + A + B which becomes eq.5d) 3×B + 2×A = 2×F
Hint #8
Substitute 2×B + A for D (from eq.6b) in eq.4a: B + 2×C = 2×B + A + A which becomes B + 2×C = 2×B + 2×A Subtract B from each side of the equation above: B + 2×C - B = 2×B + 2×A - B which becomes eq.4b) 2×C = B + 2×A
Hint #9
Multiply both sides of eq.3 by 2: 2 × (A - B) = 2 × (F - C) which becomes eq.3a) 2×A - 2×B = 2×F - 2×C
Hint #10
Substitute 3×B + 2×A for 2×F (from eq.5d) and (B + 2×A) for 2×C (from eq.4b) in eq.3a: 2×A - 2×B = 3×B + 2×A - (B + 2×A) which becomes 2×A - 2×B = 3×B + 2×A - B - 2×A which becomes 2×A - 2×B = 2×B Add 2×B to both sides of the above equation: 2×A - 2×B + 2×B = 2×B + 2×B which makes 2×A = 4×B Divide both sides by 2: 2×A ÷ 2 = 4×B ÷ 2 which makes A = 2×B
Hint #11
Substitute (2×B) for A in eq.4b: 2×C = B + 2×(2×B) which becomes 2×C = B + 4×B which makes 2×C = 5×B Divide both sides of the above equation by 2: 2×C ÷ 2 = 5×B ÷ 2 which makes C = 2½×B
Hint #12
Substitute 2½×B for C in eq.5b: B + 2½×B = F which makes 3½×B = F
Hint #13
Substitute 2×B for A in eq.6b: D = 2×B + 2×B which makes D = 4×B
Solution
Substitute 2×B for A, 2½×B for C, 4×B for D, 0 for E, and 3½×B for F in eq.1: 2×B + B + 2½×B + 4×B + 0 + 3½×B = 26 which simplifies to 13×B = 26 Divide both sides of the above equation by 13: 13×B ÷ 13 = 26 ÷ 13 which means B = 2 making A = 2×B = 2 × 2 = 4 C = 2½×B = 2½ × 2 = 5 D = 4×B = 4 × 2 = 8 F = 3½×B = 3½ × 2 = 7 and ABCDEF = 425807