Puzzle for April 15, 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 E + F with D (from eq.6): D + D = A + C which becomes eq.3a) 2×D = A + C
Hint #2
Subtract the left and right sides of eq.5 from the left and right sides of eq.2, respectively: B + F – (B + C – F) = A + D – (D + F) which becomes B + F – B – C + F = A + D – D – F which becomes 2×F – C = A – F Add C and F to both sides of the equation above: 2×F – C + C + F = A – F + C + F which becomes eq.2a) 3×F = A + C
Hint #3
In eq.2a, replace A + C with 2×D (from eq.3a): 3×F = 2×D Divide both sides of the above equation by 2: 3×F ÷ 2 = 2×D ÷ 2 which makes 1½×F = D
Hint #4
In eq.6, substitute 1½×F for D: E + F = 1½×F Subtract F from each side of the equation above: E + F – F = 1½×F – F which makes E = ½×F
Hint #5
Substitute 1½×F for D in eq.4: C + 1½×F = A + F Subtract F from each side of the above equation: C + 1½×F – F = A + F – F which becomes eq.4a) C + ½×F = A
Hint #6
Substitute C + ½×F for A (from eq.4a) in eq.2a: 3×F = C + ½×F + C which becomes 3×F = 2×C + ½×F Subtract ½×F from both sides of the above equation: 3×F – ½×F = 2×C + ½×F – ½×F which makes 2½×F = 2×C Divide both sides by 2: 2½×F ÷ 2 = 2×C ÷ 2 which makes 1¼×F = C
Hint #7
Substitute 1¼×F for C in eq.4a: 1¼×F + ½×F = A which makes 1¾×F = A
Hint #8
Substitute 1¾×F for A, and 1½×F for D in eq.2: B + F = 1¾×F + 1½×F which becomes B + F = 3¼×F Subtract F from both sides of the equation above: B + F – F = 3¼×F – F which makes B = 2¼×F
Solution
Substitute 1¾×F for A, 2¼×F for B, 1¼×F for C, 1½×F for D, and ½×F for E in eq.1: 1¾×F + 2¼×F + 1¼×F + 1½×F + ½×F + F = 33 which simplifies to 8¼×F = 33 Divide both sides of the above equation by 8¼: 8¼×F ÷ 8¼ = 33 ÷ 8¼ which means F = 4 making A = 1¾×F = 1¾ × 4 = 7 B = 2¼×F = 2¼ × 4 = 9 C = 1¼×F = 1¼ × 4 = 5 D = 1½×F = 1½ × 4 = 6 E = ½×F = ½ × 4 = 2 and ABCDEF = 795624