Puzzle for March 10, 2023 ( )
Scratchpad
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.4 may be written as: C - D + E = A + D + F + B In the above equation, replace A + D + F with E (from eq.3): C - D + E = E + B Add D to both sides, and subtract E from both sides: C - D + E + D - E = E + B + D - E which becomes eq.4a) C = B + D
Hint #2
eq.3 may be written as: E = A + F + D In the above equation, replace A + F with B (from eq.2): eq.3a) E = B + D
Hint #3
In eq.3a, substitute C for B + D (from eq.4a): E = C
Hint #4
eq.5 may be written as: B = (A + D + E + F) ÷ 4 Multiply both sides of the above equation by 4: 4 × B = 4 × (A + D + E + F) ÷ 4 which becomes 4×B = A + D + E + F which is the same as eq.5a) 4×B = A + D + F + E
Hint #5
In eq.5a, replace A + D + F with E (from eq.3): 4×B = E + E which makes 4×B = 2×E Divide both sides of the above equation by 2: 4×B ÷ 2 = 2×E ÷ 2 which makes 2×B = E and also makes eq.5b) 2×B = E = C
Hint #6
Substitute 2×B for E in eq.3a: 2×B = B + D Subtract B from each side of the above equation: 2×B - B = B + D - B which makes B = D
Hint #7
Substitute 2×B for C in eq.6: A - F = 2×B ÷ B which becomes A - F = 2 Add F to both sides of the equation above: A - F + F = 2 + F which becomes eq.6a) A = 2 + F
Hint #8
Substitute 2 + F for A (from eq.6a) into eq.2: B = 2 + F + F which makes B = 2 + 2×F and also makes eq.2a) D = B = 2 + 2×F
Hint #9
Substitute 2 + 2×F for B (from eq.2a) in eq.5b: 2×(2 + 2×F) = E = C which becomes eq.5c) 4 + 4×F = E = C
Hint #10
Substitute 2 + F for A (from eq.6a), 2 + 2×F for B and D (from eq.2a), and 4 + 4×F for C and E (from eq.5c) in eq.1: 2 + F + 2 + 2×F + 4 + 4×F + 2 + 2×F + 4 + 4×F + F = 28 which simplifies to 14 + 14×F = 28 Subtract 14 from each side of the above equation: 14 + 14×F - 14 = 28 - 14 which makes 14×F = 14 Divide both sides by 14: 14×F ÷ 14 = 14 ÷ 14 which means F = 1
Solution
Since F = 1, then: A = 2 + F = 2 + 1 = 3 (from eq.6a) B = D = 2 + 2×F = 2 + 2×1 = 2 + 2 = 4 (from eq.2a) C = E = 4 + 4×F = 4 + 4×1 = 4 + 4 = 8 (from eq.5c) and ABCDEF = 348481