Puzzle for December 24, 2022 ( )
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.
Scratchpad
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Hint #1
Subtract D and B from both sides of eq.2: C + D - D - B = A + B - D - B which becomes eq.2a) C - B = A - D Subtract D and F from both sides of eq.3: B + D - D - F = A + F - D - F which becomes eq.3a) B - F = A - D
Hint #2
In eq.3a, replace A - D with C - B (from eq.2a): B - F = C - B Add F and B to both sides of the above equation: B - F + F + B = C - B + F + B which becomes eq.3b) 2×B = C + F
Hint #3
In eq.5, replace C + F with 2×B (from eq.3b): B + E = 2×B - B which becomes B + E = B Subtract B from each side of the equation above: B + E - B = B - B which makes E = 0
Hint #4
In eq.4, substitute 0 for E: D + 0 = B + C - D which becomes D = B + C - D Add D to both sides of the equation above: D + D = B + C - D + D which becomes eq.4a) 2×D = B + C
Hint #5
In eq.6, substitute 0 for E: A + C - D = D - 0 + F which becomes A + C - D = D + F Add D to both sides of the above equation: A + C - D + D = D + F + D which becomes eq.6a) A + C = 2×D + F
Hint #6
Substitute B + C for 2×D (from eq.4a) in eq.6a: A + C = B + C + F Subtract C from each side of the above equation: A + C - C = B + C + F - C which becomes eq.6b) A = B + F
Hint #7
Substitute B + F for A (from eq.6b) into eq.3: B + D = B + F + F which becomes B + D = B + 2×F Subtract B from both sides of the equation above: B + D - B = B + 2×F - B which makes D = 2×F
Hint #8
Subtract B from both sides of eq.6b: A - B = B + F - B which becomes A - B = F Substitute A - B for F in eq.3b: 2×B = C + A - B Add B to both sides of the equation above: 2×B + B = C + A - B + B which becomes 3×B = C + A which may be written as eq.3c) 3×B = A + C
Hint #9
Substitute 3×B for A + C (from eq.3c), and (2×F) for D in eq.6a: 3×B = 2×(2×F) + F which becomes 3×B = 4×F + F which makes 3×B = 5×F Divide both sides of the above equation by 3: 3×B ÷ 3 = 5×F ÷ 3 which makes B = 1⅔×F
Hint #10
Substitute 1⅔×F for B in eq.6b: A = 1⅔×F + F which makes A = 2⅔×F
Hint #11
Substitute (1⅔×F) for B in eq.3b: 2×(1⅔×F) = C + F which becomes 3⅓×F = C + F Subtract F from both sides of the equation above: 3⅓×F - F = C + F - F which makes 2⅓×F = C
Solution
Substitute 2⅔×F for A, 1⅔×F for B, 2⅓×F for C, 2×F for D, and 0 for E in eq.1: 2⅔×F + 1⅔×F + 2⅓×F + 2×F + 0 + F = 29 which simplifies to 9⅔×F = 29 Divide both sides of the above equation by 9⅔: 9⅔×F ÷ 9⅔ = 29 ÷ 9⅔ which means F = 3 making A = 2⅔×F = 2⅔ × 3 = 8 B = 1⅔×F = 1⅔ × 3 = 5 C = 2⅓×F = 2⅓ × 3 = 7 D = 2×F = 2 × 3 = 6 and ABCDEF = 857603