Puzzle for April 13, 2022 ( )
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
Add C to both sides of eq.3: F – C + C = C + E + C which becomes eq.3a) F = 2×C + E In eq.4, replace F with 2×C + E (from eq.3a): A + D + 2×C + E = B + C Subtract C from each side of the above equation: A + D + 2×C + E – C = B + C – C which becomes eq.4a) A + D + C + E = B
Hint #2
In eq.5, replace B with A + D + C + E (from eq.4a): C + D + E = A + D + C + E – D which becomes C + D + E = A + C + E Subtract C and E from both sides of the equation above: C + D + E – C – E = A + C + E – C – E which simplifies to D = A
Hint #3
In eq.4, substitute A for D, and C + F for B (from eq.2): A + A + F = C + F + C which becomes 2×A + F = 2×C + F Subtract F from each side of the equation above: 2×A + F – F = 2×C + F – F which becomes 2×A = 2×C Divide both sides by 2: 2×A ÷ 2 = 2×C ÷ 2 which makes A = C and also makes D = A = C
Hint #4
Substitute 2×C + E for F (from eq.3a) in eq.2: B = C + 2×C + E which becomes eq.2a) B = 3×C + E
Hint #5
eq.6 may be written as: E = (B + C + F) ÷ 3 Multiply both sides of the above equation by 3: 3 × E = 3 × (B + C + F) ÷ 3 which becomes eq.6a) 3×E = B + C + F
Hint #6
Substitute 3×C + E for B (from eq.2a), and 2×C + E for F (from eq.3a) in eq.6a: 3×E = 3×C + E + C + 2×C + E which becomes 3×E = 6×C + 2×E Subtract 2×E from both sides of the equation above: 3×E – 2×E = 6×C + 2×E – 2×E which makes E = 6×C
Hint #7
Substitute 6×C for E in eq.2a: B = 3×C + 6×C which makes B = 9×C
Hint #8
Substitute 6×C for E in eq.3a: F = 2×C + 6×C which makes F = 8×C
Solution
Substitute C for A and D, 9×C for B, 6×C for E, and 8×C for F in eq.1: C + 9×C + C + C + 6×C + 8×C = 26 which simplifies to 26×C = 26 Divide both sides of the above equation by 26: 26×C ÷ 26 = 26 ÷ 26 which means C = 1 making A = D = C = 1 B = 9×C = 9×1 = 9 E = 6×C = 6×1 = 6 F = 8×C = 8×1 = 8 and ABCDEF = 191168