Puzzle for October 29, 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.
* BC and CD are 2-digit numbers (not B×C or C×D).
Scratchpad
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Hint #1
In eq.3, replace F with B - D (from eq.4): E = C + B - D Add D to both sides of the above equation: E + D = C + B - D + D which becomes eq.3a) E + D = C + B
Hint #2
eq.6 may be written as: C = (A + B + D + E) ÷ 4 Multiply both sides of the above equation by 4: 4 × C = 4 × (A + B + D + E) ÷ 4 which becomes 4×C = A + B + D + E which may be written as eq.6a) 4×C = A + B + E + D
Hint #3
In eq.6a, replace E + D with C + B (from eq.3a): 4×C = A + B + C + B which becomes 4×C = A + 2×B + C Subtract C from each side of the above equation: 4×C - C = A + 2×B + C - C which becomes eq.6b) 3×C = A + 2×B
Hint #4
In eq.6b, substitute (A + C) for B (from eq.2): 3×C = A + 2×(A + C) which becomes 3×C = A + 2×A + 2×C which becomes 3×C = 3×A + 2×C Subtract 2×C from each side of the equation above: 3×C - 2×C = 3×A + 2×C - 2×C which makes C = 3×A
Hint #5
Substitute 3×A for C in eq.2: B = A + 3×A which makes B = 4×A
Hint #6
Substitute (3×A) for C, and 4×A for B in eq.6a: 4×(3×A) = A + 4×A + E + D which becomes 12×A = 5×A + E + D Subtract 5×A from both sides of the equation above: 12×A - 5×A = 5×A + E + D - 5×A which becomes eq.6c) 7×A = E + D
Hint #7
eq.5 may be written as: A + 10×B + C - (10×C + D + E) = C + E + F which becomes A + 10×B + C - 10×C - D - E = C + E + F which becomes A + 10×B - 9×C - D - E = C + E + F Add 9×C, D, and E to both sides of the above equation: A + 10×B - 9×C - D - E + 9×C + D + E = C + E + F + 9×C + D + E which becomes A + 10×B = 10×C + 2×E + F + D which may be written as eq.5a) A + 10×B = 10×C + E + E + D + F
Hint #8
Substitute (4×A) for B, (3×A) for C, and 7×A for E + D (from eq.6c) in eq.5a: A + 10×(4×A) = 10×(3×A) + E + 7×A + F which becomes A + 40×A = 30×A + 7×A + E + F which becomes 41×A = 37×A + E + F Subtract 37×A from each side of the equation above: 41×A - 37×A = 37×A + E + F - 37×A which becomes eq.5b) 4×A = E + F
Hint #9
Substitute B for 4×A in eq.5b: eq.5c) B = E + F
Hint #10
Substitute E + F for B (from eq.5c) into eq.4: F = E + F - D In the equation above, subtract F from both sides, and add D to both sides: F - F + D = E + F - D - F + D which simplifies to D = E
Hint #11
Substitute D for E in eq.6c: 7×A = D + D which makes 7×A = 2×D Divide both sides of the above equation by 2: 7×A ÷ 2 = 2×D ÷ 2 which makes 3½×A = D and also makes 3½×A = D = E
Hint #12
Substitute 3½×A for E in eq.5b: 4×A = 3½×A + F Subtract 3½×A from each side of the equation above: 4×A = 3½×A + F which makes ½×A = F
Solution
Substitute 4×A for B, 3×A for C, 3½×A for D and E, and ½×A for F in eq.1: A + 4×A + 3×A + 3½×A + 3½×A + ½×A = 31 which simplifies to 15½×A = 31 Divide both sides of the above equation by 15½: 15½×A ÷ 15½ = 31 ÷ 15½ which means A = 2 making B = 4×A = 4 × 2 = 8 C = 3×A = 3 × 2 = 6 D = E = 3½×A = 3½ × 2 = 7 F = ½×A = ½ × 2 = 1 and ABCDEF = 286771