Puzzle for November 8, 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
Add A and E to both sides of eq.6: C – A + A + E = D – E + F + A + E which becomes C + E = D + F + A In the above equation, replace D + F with A + C (from eq.3): C + E = A + C + A Subtract C from both sides: C + E – C = A + C + A – C which makes E = 2×A
Hint #2
Add E to both sides of eq.5: B + C + D – E + E = A + E + E which becomes B + C + D = A + 2×E In eq.1, replace B + C + D with A + 2×E: A + A + 2×E + E + F = 28 which becomes eq.1a) 2×A + 3×E + F = 28
Hint #3
In eq.1a, substitute (2×A) for E: 2×A + 3×(2×A) + F = 28 which is equivalent to 2×A + 6×A + F = 28 which becomes 8×A + F = 28 Subtract 8×A from both sides of the above equation: 8×A + F – 8×A = 28 – 8×A which makes eq.1b) F = 28 – 8×A
Hint #4
Substitute 2×A for E, and 28 – 8×A for F (from eq.1b) in eq.2: 2×A + 28 – 8×A = A + D which becomes 28 – 6×A = A + D Subtract A from each side of the equation above: 28 – 6×A – A = A + D – A which makes eq.2a) 28 – 7×A = D
Hint #5
Substitute 28 – 7×A for D (from eq.2a), and 28 – 8×A for F (from eq.1b) in eq.3: 28 – 7×A + 28 – 8×A = A + C which becomes 56 – 15×A = A + C Subtract A from each side of the above equation: 56 – 15×A – A = A + C – A which makes eq.3a) 56 – 16×A = C
Hint #6
Substitute 56 – 16×A for C (from eq.3a), 28 – 7×A for D (from eq.2a), and 2×A for E in eq.5: B + 56 – 16×A + 28 – 7×A – 2×A = A + 2×A which becomes B + 84 – 25×A = 3×A In the above equation, add 25×A to both sides, and subtract 84 from each side: B + 84 – 25×A + 25×A – 84 = 3×A + 25×A – 84 which makes eq.5a) B = 28×A – 84
Solution
Substitute 28 – 8×A for F (from eq.1b), 28×A – 84 for B (from eq.5a), and 28 – 7×A for D (from eq.2a) in eq.4: A + 28 – 8×A = 28×A – 84 + 28 – 7×A which becomes 28 – 7×A = 21×A – 56 Add 7×A and 56 to both sides of the above equation: 28 – 7×A + 7×A + 56 = 21×A – 56 + 7×A + 56 which makes 84 = 28×A Divide both sides by 28: 84 ÷ 28 = 28×A ÷ 28 which means 3 = A making B = 28×A – 84 = 28×3 – 84 = 84 – 84 = 0 (from eq.5a) C = 56 – 16×A = 56 – 16×3 = 56 – 48 = 8 (from eq.3a) D = 28 – 7×A = 28 – 7×3 = 28 – 21 = 7 (from eq.2a) E = 2×A = 2×3 = 6 F = 28 – 8×A = 28 – 8×3 = 28 – 24 = 4 (from eq.1b) and ABCDEF = 308764