Puzzle for November 27, 2021 ( )
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 E with B + C (from eq.2): B + C + F = B + D Subtract B from both sides of the equation above: B + C + F – B = B + D – B which becomes eq.3a) C + F = D
Hint #2
In eq.4, replace B + C with E (from eq.2): E + F = A – C + E In the above equation, subtract E from both sides, and add C to both sides: E + F – E + C = A – C + E – E + C which becomes F + C = A which may be written as eq.4a) C + F = A
Hint #3
eq.6 may be written as: F = (A + C + E) ÷ 3 Multiply both sides of the above equation by 3: 3 × F = 3 × (A + C + E) ÷ 3 which becomes eq.6a) 3×F = A + C + E
Hint #4
Substitute C + F for A (from eq.4a) in eq.6a: 3×F = C + F + C + E which becomes 3×F = 2×C + F + E Subtract 2×C and F from each side of the equation above: 3×F – 2×C – F = F + 2×C + E – 2×C – F which becomes eq.6b) 2×F – 2×C = E
Hint #5
Subtract C from each side of eq.2: B + C – C = E – C which becomes B = E – C Substitute 2×F – 2×C for E (from eq.6b) into the above equation: B = 2×F – 2×C – C which becomes eq.2a) B = 2×F – 3×C
Hint #6
Substitute C + F for D (from eq.3a), (2×F – 2×C) for E (from eq.6b), and 2×F – 3×C for B (from eq.2a) in eq.5: C + F – (2×F – 2×C) + F = 2×F – 3×C + C – F which becomes C + 2×F – 2×F + 2×C = F – 2×C which becomes 3×C = F – 2×C Add 2×C to both sides of the above equation: 3×C + 2×C = F – 2×C + 2×C which makes 5×C = F
Hint #7
Substitute (5×C) for F in eq.6b: 2×(5×C) – 2×C = E which becomes 10×C – 2×C = E which makes 8×C = E
Hint #8
Substitute (5×C) for F in eq.2a: B = 2×(5×C) – 3×C which becomes B = 10×C – 3×C which makes B = 7×C
Hint #9
Substitute 5×C for F in eq.4a: C + 5×C = A which makes 6×C = A
Hint #10
Substitute 5×C for F in eq.3a: C + 5×C = D which makes 6×C = D
Solution
Substitute 6×C for A and D, 7×C for B, 8×C for E, and 5×C for F in eq.1: 6×C + 7×C + C + 6×C + 8×C + 5×C = 33 which simplifies to 33×C = 33 Divide both sides of the above equation by 33: 33×C ÷ 33 = 33 ÷ 33 which means C = 1 making A = D = 6×C = 6×1 = 6 B = 7×C = 7×1 = 7 E = 8×C = 8×1 = 8 F = 5×C = 5×1 = 5 and ABCDEF = 671685