Puzzle for May 30, 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.2, replace B + C with A + F (from eq.4): D + F = A + F + E Subtract F from both sides of the above equation: D + F - F = A + F + E - F which makes eq.2a) D = A + E
Hint #2
In eq.3, replace D with A + E (from eq.2a): C + A + E + F = A + B Subtract A from both sides of the above equation: C + A + E + F - A = A + B - A which becomes eq.3a) C + E + F = B
Hint #3
In eq.6, substitute C + E + F for B (from eq.3a): C + E + F = C + D + E Subtract both C and E from each side of the above equation: C + E + F - C - E = C + D + E - C - E which makes F = D
Hint #4
Add A to both sides of eq.5: A + C + A = B - A + A which becomes eq.5a) 2×A + C = B Substitute D for F, and 2×A + C for B (from eq.5a) in eq.3: C + D + D = A + 2×A + C which becomes C + 2×D = 3×A + C Subtract C from both sides of the above equation: C + 2×D - C = 3×A + C - C which makes 2×D = 3×A Divide each side by 2: 2×D ÷ 2 = 3×A ÷ 2 which makes D = 1½×A which also makes F = D = 1½×A
Hint #5
Substitute 1½×A for D in eq.2a: 1½×A = A + E Subtract A from each side of the equation above: 1½×A - A = A + E - A which makes ½×A = E
Hint #6
Substitute 1½×A for F, and 2×A + C for B (from eq.5a) in eq.4: 2×A + C + C = A + 1½×A which becomes 2×A + 2×C = 2½×A Subtract 2×A from both sides: 2×A + 2×C - 2×A = 2½×A - 2×A which makes 2×C = ½×A Divide both sides by 2: 2×C ÷ 2 = ½×A ÷ 2 which means C = ¼×A
Hint #7
Substitute ¼×A for C in eq.5a: 2×A + ¼×A = B which makes 2¼×A = B
Solution
Substitute 2¼×A for B, ¼×A for C, 1½×A for D and F, and ½×A for E in eq.1: A + 2¼×A + ¼×A + 1½×A + ½×A + 1½×A = 28 which simplifies to 7×A = 28 Divide both sides of the above equation by 7: 7×A ÷ 7 = 28 ÷ 7 which means A = 4 making B = 2¼×A = 2¼ × 4 = 9 C = ¼×A = ¼ × 4 = 1 D = F = 1½×A = 1½ × 4 = 6 E = ½×A = ½ × 4 = 2 and ABCDEF = 491626