Puzzle for August 27, 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.5, replace B + C with E + F (from eq.2): E + F = A - D + F In the above equation, subtract F from both sides, and add D to both sides: E + F - F + D = A - D + F - F + D which becomes E + D = A which is the same as eq.5a) D + E = A
Hint #2
In eq.4, replace D + E with A (from eq.5a): eq.4a) A = C + F
Hint #3
In eq.5, substitute C + F for A (from eq.4a): B + C = C + F - D + F which becomes B + C = C + 2×F - D In the above equation, subtract C from both sides, and add D to both sides: B + C - C + D = C + 2×F - D - C + D which becomes eq.5b) B + D = 2×F
Hint #4
Add A and F to both sides of eq.3: F - A + A + F = B - F + A + F which becomes eq.3a) 2×F = B + A
Hint #5
Substitute B + A for 2×F (from eq.3a) into eq.5b: B + D = B + A Subtract B from both sides of the above equation: B + D - B = B + A - B which makes D = A
Hint #6
Substitute A for D in eq.5a: A + E = A Subtract A from each side of the equation above: A + E - A = A - A which makes E = 0
Hint #7
eq.6 may be written as: C + F - B = (A + C + D + F) ÷ 4 Multiply both sides of the above equation by 4: 4 × (C + F - B) = 4 × (A + C + D + F) ÷ 4 which becomes eq.6a) 4×C + 4×F - 4×B = A + C + D + F
Hint #8
Substitute A for D in eq.6a: 4×C + 4×F - 4×B = A + C + A + F which becomes 4×C + 4×F - 4×B = 2×A + C + F In the equation above, add 4×B to both sides, and subtract C and F from both sides: 4×C + 4×F - 4×B + 4×B - C - F = 2×A + C + F + 4×B - C - F which becomes eq.6b) 3×C + 3×F = 2×A + 4×B
Hint #9
Substitute A for D in eq.5: B + C = A - A + F which becomes eq.5c) B + C = F
Hint #10
Substitute B + C for F (from eq.5c) into eq.3a: 2×(B + C) = B + A which becomes 2×B + 2×C = B + A Subtract B from each side of the above equation: 2×B + 2×C - B = B + A - B which becomes eq.3b) B + 2×C = A
Hint #11
Substitute B + C for F (from eq.5c) and B + 2×C for A (from eq.3b) into eq.6b: 3×C + 3×(B + C) = 2×(B + 2×C) + 4×B which becomes 3×C + 3×B + 3×C = 2×B + 4×C + 4×B which becomes 6×C + 3×B = 6×B + 4×C Subtract 3×B and 4×C from both sides of the above equation: 6×C + 3×B - 3×B - 4×C = 6×B + 4×C - 3×B - 4×C which makes 2×C = 3×B Divide both sides by 2: 2×C ÷ 2 = 3×B ÷ 2 which makes C = 1½×B
Hint #12
Substitute (1½×B) for C in eq.3b: B + 2×(1½×B) = A which becomes B + 3×B = A which makes 4×B = A and also makes 4×B = A = D
Hint #13
Substitute 1½×B for C in eq.5c: B + 1½×B = F which makes 2½×B = F
Solution
Substitute 4×B for A and D, 1½×B for C, 0 for E, and 2½×B for F in eq.1: 4×B + B + 1½×B + 4×B + 0 + 2½×B = 26 which simplifies to 13×B = 26 Divide both of the above equation sides by 13: 13×B ÷ 13 = 26 ÷ 13 which means B = 2 making A = D = 4×B = 4 × 2 = 8 C = 1½×B = 1½ × 2 = 3 F = 2½×B = 2½ × 2 = 5 and ABCDEF = 823805