Puzzle for December 23, 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
In eq.3, replace B with C + E (from eq.2): A + E = C + E + C which becomes A + E = 2×C + E Subtract E from each side of the equation above: A + E - E = 2×C + E - E which makes A = 2×C
Hint #2
eq.6 may be written as: F = (B + C + D + E) ÷ 4 Multiply both sides of the above equation by 4: 4 × F = 4 × (B + C + D + E) ÷ 4 which becomes eq.6a) 4×F = B + C + D + E
Hint #3
In eq.6a, replace C + D with E + F (from eq.4): 4×F = B + E + F + E Subtract F from each side of the above equation: 4×F - F = B + E + F + E - F which becomes eq.6b) 3×F = B + E + E
Hint #4
In eq.6b, substitute A + F for B + E (from eq.5): 3×F = A + F + E Subtract F from both sides of the above equation: 3×F - F = A + F + E - F which becomes eq.6c) 2×F = A + E
Hint #5
Substitute A + E for B + C (from eq.3) in eq.6a: eq.6d) 4×F = A + E + D + E
Hint #6
Substitute 2×F for A + E (from eq.6c) in eq.6d: 4×F = 2×F + D + E Subtract 2×F from both sides of the equation above: 4×F - 2×F = 2×F + D + E - 2×F which becomes eq.6e) 2×F = D + E
Hint #7
Substitute D + E for 2×F (from eq.6e), and 2×C for A in eq.6c: D + E = 2×C + E Subtract E from each side of the equation above: D + E - E = 2×C + E - E which makes D = 2×C
Hint #8
eq.6a may be written as: 4×F = B + E + C + D Substitute A + F for B + E (from eq.5) in the above equation: 4×F = A + F + C + D Subtract F from each side of the above equation: 4×F - F = A + F + C + D - F which becomes eq.6f) 3×F = A + C + D
Hint #9
Substitute 2×C for A and D in eq.6f: 3×F = 2×C + C + 2×C which becomes 3×F = 5×C Divide both sides of above equation by 3: 3×F ÷ 3 = 5×C ÷ 3 which makes F = 1⅔×C
Hint #10
Substitute (1⅔×C) for F, and 2×C for D in eq.6e: 2×(1⅔×C) = 2×C + E which becomes 3⅓×C = 2×C + E Subtract 2×C from both sides of the above equation: 3⅓×C - 2×C = 2×C + E - 2×C which makes 1⅓×C = E
Hint #11
Substitute (1⅔×C) for F, and 1⅓×C for E in eq.6b: 3×(1⅔×C) = B + 1⅓×C + 1⅓×C which becomes 5×C = B + 2⅔×C Subtract 2⅔×C from both sides of the equation above: 5×C - 2⅔×C = B + 2⅔×C - 2⅔×C which makes 2⅓×C = B
Solution
Substitute 2×C for A and D, 2⅓×C for B, 1⅓×C for E, and 1⅔×C for F in eq.1: 2×C + 2⅓×C + C + 2×C + 1⅓×C + 1⅔×C = 31 which simplifies to 10⅓×C = 31 Divide both sides of the above equation by 10⅓: 10⅓×C ÷ 10⅓ = 31 ÷ 10⅓ which means C = 3 making A = D = 2×C = 2 × 3 = 6 B = 2⅓×C = 2⅓ × 3 = 7 E = 1⅓×C = 1⅓ × 3 = 4 F = 1⅔×C = 1⅔ × 3 = 5 and ABCDEF = 673645