Puzzle for December 3, 2019 ( )
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, replace A with B + D (from eq.2): B – D = B + D – C Subtract B from both sides of the above equation: B – D – B = B + D – C – B which becomes –D = D – C Add C and D to each side: –D + C + D = D – C + C + D which makes eq.6a) 2×D = C
Hint #2
In eq.4, replace C with 2×D: D + F = 2×D Subtract D from both sides of the equation above: D + F – D = 2×D – D which makes F = D
Hint #3
In eq.5, substitute B + D for A (from eq.2), D for F, and 2×D for C: B + D + D = 2×D + E which becomes B + 2×D = 2×D + E Subtract 2×D from each side of the equation above: B + 2×D – 2×D = 2×D + E – 2×D which makes B = E
Hint #4
Substitute B for E in eq.3: C – B = B + B which becomes C – B = 2×B Add B to both sides of the equation above: C – B + B = 2×B + B which makes C = 3×B
Hint #5
Substitute 3×B for C in eq.6a: 2×D = 3×B Divide both sides of the equation above by 2: 2×D ÷ 2 = 3×B ÷ 2 which makes D = 1½×B and also makes F = D = 1½×B
Hint #6
Substitute 1½×B for D in eq.2: B + 1½×B = A which makes 2½×B = A
Solution
Substitute 2½×B for A, 3×B for C, 1½×B for D and F, and B for E in eq.1: 2½×B + B + 3×B + 1½×B + B + 1½×B = 21 which simplifies to 10½×B = 21 Divide both sides of the above equation by 10½: 10½×B ÷ 10½ = 21 ÷ 10½ which means B = 2 making A = 2½×B = 2½ × 2 = 5 C = 3×B = 3 × 2 = 6 D = F = 1½×B = 1½ × 2 = 3 E = B = 2 and ABCDEF = 526323