Puzzle for December 8, 2021 ( )
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.3, replace A with C + E (from eq.2): B + E = C + E + C which becomes B + E = 2×C + E Subtract E from each side of the above equation: B + E – E = 2×C + E – E which makes eq.3a) B = 2×C
Hint #2
In eq.6, replace B + C with A + F (from eq.4): D + F = A + A + F which becomes D + F = 2×A + F Subtract F from each side of the above equation: D + F – F = 2×A + F – F which makes D = 2×A
Hint #3
In eq.4, substitute 2×C for B (from eq.3a): A + F = 2×C + C which becomes eq.4a) A + F = 3×C
Hint #4
In eq.5, substitute 2×A for D, and 2×C for B (from eq.3a): C + 2×A = 2×C + F Subtract 2×C from each side of the equation above: C + 2×A – 2×C = 2×C + F – 2×C which becomes eq.5a) 2×A – C = F
Hint #5
Substitute 2×A – C for F (from eq.5a) into eq.4a: A + 2×A – C = 3×C which becomes 3×A – C = 3×C Add C to both sides of the above equation: 3×A – C + C = 3×C + C which makes 3×A = 4×C Divide both sides by 4: 3×A ÷ 4 = 4×C ÷ 4 which makes ¾×A = C
Hint #6
Substitute (¾×A) for C in eq.3a: B = 2×(¾×A) which makes B = 1½×A
Hint #7
Substitute ¾×A for C in eq.5a: 2×A – ¾×A = F which makes 1¼×A = F
Hint #8
Substitute ¾×A for C in eq.2: ¾×A + E = A Subtract ¾×A from each side of the equation above: ¾×A + E – ¾×A = A – ¾×A which makes E = ¼×A
Solution
Substitute 1½×A for B, ¾×A for C, 2×A for D, ¼×A for E, and 1¼×A for F in eq.1: A + 1½×A + ¾×A + 2×A + ¼×A + 1¼×A = 27 which simplifies to 6¾×A = 27 Divide both sides of the above equation by 6¾: 6¾×A ÷ 6¾ = 27 ÷ 6¾ which means A = 4 making B = 1½×A = 1½ × 4 = 6 C = ¾×A = ¾ × 4 = 3 D = 2×A = 2 × 4 = 8 E = ¼×A = ¼ × 4 = 1 F = 1¼×A = 1¼ × 4 = 5 and ABCDEF = 463815