Puzzle for December 4, 2020 ( )
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.6, substitute (F ÷ B) for A (from eq.5): (F ÷ B) ÷ F = E ÷ B which may be written as (F / B) × (1 / F) = E ÷ B which becomes 1 ÷ B = E ÷ B Multiply both sides of the above equation by B: (1 ÷ B) × B = (E ÷ B) × B which makes 1 = E
Hint #2
In eq.4, replace E with 1: A – D = B × 1 which is equivalent to A – D = B Add D to both sides of the equation above: A – D + D = B + D which becomes eq.4a) A = B + D
Hint #3
In eq.1, replace B + D with A (from eq.4a): A = A + C Subtract A from each side of the equation above: A – A = A + C – A which makes 0 = C
Hint #4
In eq.2, substitute 1 for E, and 0 for C: 1 + F = A + B + 0 which becomes 1 + F = A + B Subtract 1 from each side of the equation above: 1 + F – 1 = A + B – 1 which becomes eq.2a) F = A + B – 1
Hint #5
Substitute 1 for E, 0 for C, and A + B – 1 for F (from eq.2a) in eq.3: B + 1 + A + B – 1 = A + 0 + D – 1 which becomes A + 2×B = A + D – 1 In the equation above, add 1 to each side, and subtract A from each side: A + 2×B + 1 – A = A + D – 1 + 1 – A which becomes eq.3a) 2×B + 1 = D
Hint #6
Substitute 2×B + 1 for D (from eq.3a) in eq.4a: A = B + 2×B + 1 which becomes eq.4b) A = 3×B + 1
Hint #7
Substitute 3×B + 1 for A (from eq.4b) in eq.2a: F = 3×B + 1 + B – 1 which makes F = 4×B
Solution
Substitute 4×B for F, and 3×B + 1 for A (from eq.4b) in eq.5: 4×B ÷ B = 3×B + 1 which becomes 4 = 3×B + 1 Subtract 1 from both sides of the above equation: 4 – 1 = 3×B + 1 – 1 which makes 3 = 3×B Divide both sides by 3: 3 ÷ 3 = 3×B ÷ 3 which means 1 = B making A = 3×B + 1 = 3×1 + 1 = 3 + 1 = 4 (from eq.4b) D = 2×B + 1 = 2×1 + 1 = 2 + 1 = 3 (from eq.3a) F = 4×B = 4×1 = 4 and ABCDEF = 410314