Puzzle for April 10, 2020 ( )
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
eq.2 may be written as: D + F = B + E + C In the above equation, replace B + E with A + C (from eq.3): D + F = A + C + C which becomes D + F = A + 2×C Subtract 2×C from both sides of the above equation: D + F – 2×C = A + 2×C – 2×C which becomes eq.2a) D + F – 2×C = A
Hint #2
In eq.6, replace A with D + F – 2×C (from eq.2a): D + F – 2×C – C – F = C – D + F which becomes D – 3×C = C – D + F In the equation above, add D to both sides, and subtract C from both sides: D – 3×C + D – C = C – D + F + D – C which becomes eq.6a) 2×D – 4×C = F
Hint #3
Substitute 2×D – 4×C for F (from eq.6a) in eq.2a: D + 2×D – 4×C – 2×C = A which becomes eq.2b) 3×D – 6×C = A
Hint #4
Add A and F to both sides of eq.5: C – A + A + F = A – F + A + F which becomes C + F = 2×A Substitute (3×D – 6×C) for A (from eq.2b) in the above equation: C + F = 2×(3×D – 6×C) which becomes C + F = 6×D – 12×C Subtract C from both sides: C + F – C = 6×D – 12×C – C which becomes eq.5a) F = 6×D – 13×C
Hint #5
Substitute 6×D – 13×C for F (from eq.5a) in eq.6a: 2×D – 4×C = 6×D – 13×C In the above equation, subtract 2×D from both sides, and add 13×C to both sides: 2×D – 4×C – 2×D + 13×C = 6×D – 13×C – 2×D + 13×C which simplifies to 9×C = 4×D Divide both sides by 4: 9×C ÷ 4 = 4×D ÷ 4 which makes 2¼×C = D
Hint #6
Substitute (2¼×C) for D in eq.6a: 2×(2¼×C) – 4×C = F which becomes 4½×C – 4×C = F which makes ½×C = F
Hint #7
Substitute (2¼×C) for D, and ½×C for F in eq.2b: 3×(2¼×C) – 6×C = A which becomes 6¾×C – 6×C = A which makes ¾×C = A
Hint #8
Substitute ¾×C for A, ½×C for F, and 2¼×C for D in eq.4: ¾×C + C + ½×C = B + 2¼×C which becomes 2¼×C = B + 2¼×C Subtract 2¼×C from both sides of the equation above: 2¼×C – 2¼×C = B + 2¼×C – 2¼×C which makes 0 = B
Hint #9
Substitute 0 for B, and ¾×C for A in eq.3: 0 + E = ¾×C + C which makes E = 1¾×C
Solution
Substitute ¾×C for A, 0 for B, 2¼×C for D, 1¾×C for E, and ½×C for F in eq.1: ¾×C + 0 + C + 2¼×C + 1¾×C + ½×C = 25 which simplifies to 6¼×C = 25 Divide both sides of the above equation by 6¼: 6¼×C ÷ 6¼ = 25 ÷ 6¼ which means C = 4 making A = ¾×C = ¾ × 4 = 3 D = 2¼×C = 2¼ × 4 = 9 E = 1¾×C = 1¾ × 4 = 7 F = ½×C = ½ × 4 = 2 and ABCDEF = 304972