Puzzle for December 30, 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.
Scratchpad
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Hint #1
eq.6 may be written as: B - C = (A + D + E + F) ÷ 4 Multiply both sides of the above equation by 4: 4 × (B - C) = 4 × (A + D + E + F) ÷ 4 which becomes eq.6a) 4×B - 4×C = A + D + E + F
Hint #2
In eq.6a, replace D + E + F with B + C (from eq.5): 4×B - 4×C = A + B + C Subtract B and C from both sides of the equation above: 4×B - 4×C - B - C = A + B + C - B - C which becomes eq.6b) 3×B - 5×C = A
Hint #3
In eq.6b, substitute B + E for A (from eq.2): 3×B - 5×C = B + E Subtract B from each side of the equation above: 3×B - 5×C - B = B + E - B which becomes eq.6c) 2×B - 5×C = E
Hint #4
Substitute (2×B - 5×C) for E (from eq.6c) in eq.4: (2×B - 5×C) + F = C - (2×B - 5×C) which becomes 2×B - 5×C + F = C - 2×B + 5×C which becomes 2×B - 5×C + F = 6×C - 2×B In the equation above, subtract 2×B from both sides, and add 5×C to both sides: 2×B - 5×C + F - 2×B + 5×C = 6×C - 2×B - 2×B + 5×C which becomes eq.4a) F = 11×C - 4×B
Hint #5
Substitute A + F for D + E (from eq.3) in eq.6a: 4×B - 4×C = A + A + F + F which becomes 4×B - 4×C = 2×A + 2×F Divide both sides of the above equation by 2: (4×B - 4×C) ÷ 2 = (2×A + 2×F) ÷ 2 which becomes eq.6d) 2×B - 2×C = A + F
Hint #6
Substitute 3×B - 5×C for A (from eq.6b), and 11×C - 4×B for F (from eq.4a) in eq.6d: 2×B - 2×C = 3×B - 5×C + 11×C - 4×B which becomes 2×B - 2×C = -B + 6×C Add 2×C and B to both sides of the equation above: 2×B - 2×C + 2×C + B = -B + 6×C + 2×C + B which makes 3×B = 8×C Divide both sides by 8: 3×B ÷ 8 = 8×C ÷ 8 which makes eq.6e) ⅜×B = C
Hint #7
Substitute (⅜×B) for C in eq.4a: F = 11×(⅜×B) - 4×B which becomes F = 4⅛×B - 4×B which makes F = ⅛×B Multiply both sides by 8: 8 × F = 8 × ⅛×B which makes 8×F = B
Hint #8
Substitute (8×F) for B in eq.6e: ⅜×(8×F) = C which makes 3×F = C
Hint #9
Substitute (8×F) for B, and (3×F) for C in eq.6c: 2×(8×F) - 5×(3×F) = E which becomes 16×F - 15×F = E which makes F = E
Hint #10
Substitute (8×F) for B, (3×F) for C in eq.6b: 3×(8×F) - 5×(3×F) = A which becomes 24×F - 15×F = A which makes 9×F = A
Hint #11
Substitute F for E, and 9×F for A in eq.3: D + F = 9×F + F which becomes D + F = 10×F Subtract F from each side of the above equation: D + F - F = 10×F - F which makes D = 9×F
Solution
Substitute 9×F for A and D, 8×F for B, 3×F for C, and F for E in eq.1: 9×F + 8×F + 3×F + 9×F + F + F = 31 which simplifies to 31×F = 31 Divide both sides of the above equation by 31: 31×F ÷ 31 = 31 ÷ 31 which means F = 1 making A = D = 9×F = 9 × 1 = 9 B = 8×F = 8 × 1 = 8 C = 3×F = 3 × 1 = 3 E = F = 1 and ABCDEF = 983911