Puzzle for June 10, 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: C + D + C = E – C + C which becomes eq.3a) 2×C + D = E In eq.5, replace E with 2×C + D (from eq.3a): B + C + F = D + 2×C + D – C which becomes B + C + F = 2×D + C Subtract C from each side of the equation above: B + C + F – C = 2×D + C – C which becomes eq.5a) B + F = 2×D
Hint #2
Add A and D to both sides of eq.4: D – A + A + D = B – D + A + D which becomes eq.4a) 2×D = B + A In eq.5a, replace 2×D with B + A (from eq.4a): B + F = B + A Subtract B from each side of the above equation: B + F – B = B + A – B which makes F = A
Hint #3
In eq.2, substitute A for F: A = D – A Add A to both sides of the equation above: A + A = D – A + A which makes eq.2a) 2×A = D
Hint #4
Substitute (2×A) for D (from eq.2a) in eq.4a: 2×(2×A) = B + A which becomes 4×A = B + A Subtract A from both sides of the equation above: 4×A – A = B + A – A which makes eq.4b) 3×A = B
Hint #5
eq.6 may be written as: D = (B + C + E) ÷ 3 Multiply both sides of the above equation by 3: 3 × D = 3 × (B + C + E) ÷ 3 which becomes eq.6a) 3×D = B + C + E
Hint #6
Substitute (2×A) for D (from eq.2a), and 3×A for B (from eq.4b) in eq.6a: 3×(2×A) = 3×A + C + E which becomes 6×A = 3×A + C + E Subtract 3×A and C from each side of the above equation: 6×A – 3×A – C = 3×A + C + E – 3×A – C which becomes eq.6b) 3×A – C = E
Hint #7
Substitute 2×A for D (from eq.2a), and 3×A – C for E (from eq.6b) in eq.3a: 2×C + 2×A = 3×A – C In the above equation, subtract 2×A from both sides, and add C to both sides: 2×C + 2×A – 2×A + C = 3×A – C – 2×A + C which makes 3×C = A and also makes 3×C = A = F
Hint #8
Substitute (3×C) for A in eq.6b: 3×(3×C) – C = E which becomes 9×C – C = E which makes 8×C = E
Hint #9
Substitute 3×C for A in eq.4b: 3×(3×C) = B which makes 9×C = B
Hint #10
Substitute 3×C for A in eq.2a: 2×(3×C) = D which makes 6×C = D
Solution
Substitute 3×C for A and F, 9×C for B, 6×C for D, and 8×C for E in eq.1: 3×C + 9×C + C + 6×C + 8×C + 3×C = 30 which simplifies to 30×C = 30 Divide both sides of the above equation by 30: 30×C ÷ 30 = 30 ÷ 30 which means C = 1 making A = F = 3×C = 3 × 1 = 3 B = 9×C = 9 × 1 = 9 D = 6×C = 6 × 1 = 6 E = 8×C = 8 × 1 = 8 and ABCDEF = 391683