Puzzle for July 5, 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.
* AB and DE are 2-digit numbers (not A×B or D×E).
Scratchpad
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Hint #1
Add B to both sides of eq.5: B + F + B = A – B + C + B which becomes eq.5a) 2×B + F = A + C In eq.3, replace A + C with 2×B + F (from eq.5a): B + D = 2×B + F – D + F which becomes eq.5b) B + D = 2×B + 2×F – D
Hint #2
Subtract B and 2×F from both sides of eq.5b: B + D – B – 2×F = 2×B + 2×F – D – B – 2×F which becomes D – 2×F = B – D Add D to both sides of the above equation: D – 2×F + D = B – D + D which becomes eq.5c) 2×D – 2×F = B
Hint #3
Add D and F to both sides of eq.4: D + E – F + D + F = B + C – D + D + F which becomes 2×D + E = B + C + F In the above equation, replace B with 2×D – 2×F (from eq.5c): 2×D + E = 2×D – 2×F + C + F which becomes 2×D + E = 2×D – F + C Subtract 2×D from both sides: 2×D + E – 2×D = 2×D – F + C – 2×D which becomes eq.4a) E = C – F
Hint #4
In eq.2, substitute C – F for E (from eq.4a): C – F = A – C Add C to both sides of the equation above: C – F + C = A – C + C which becomes eq.2a) 2×C – F = A
Hint #5
Substitute 2×C – F for A (from eq.2a) in eq.5a: 2×B + F = 2×C – F + C which becomes 2×B + F = 3×C – F Subtract F from each side of the equation above: 2×B + F – F = 3×C – F – F which becomes 2×B = 3×C – 2×F Divide both sides by 2: 2×B ÷ 2 = (3×C – 2×F) ÷ 2 which becomes eq.5d) B = 1½×C – F
Hint #6
Substitute 2×C – F for A (from eq.2a) in eq.3: B + D = 2×C – F + C – D + F which becomes B + D = 3×C – D Add D to both sides of the equation above: B + D + D = 3×C – D + D which becomes eq.3b) B + 2×D = 3×C
Hint #7
Substitute 1½×C – F for B (from eq.5d) in eq.3b: 1½×C – F + 2×D = 3×C In the above equation, subtract 1½×C from both sides, and add F to both sides: 1½×C – F + 2×D – 1½×C + F = 3×C – 1½×C + F which becomes 2×D = 1½×C + F Divide both sides by 2: 2×D ÷ 2 = (1½×C + F) ÷ 2 which becomes eq.3c) D = ¾×C + ½×F
Hint #8
eq.6 may be written as: 10×A + B + C = 10×D + E – F Substitute (2×C – F) for A (from eq.2a), 1½×C – F for B (from eq.5d), (¾×C + ½×F) for D (from eq.3c), and C – F for E (from eq.4a) in the above equation: 10×(2×C – F) + 1½×C – F + C = 10×(¾×C + ½×F) + C – F – F which becomes 20×C – 10×F + 2½×C – F = 7½×C + 5×F + C – 2×F which becomes 22½×C – 11×F = 8½×C + 3×F In the equation above, add 11×F to each side, and subtract 8½×C from each side: 22½×C – 11×F + 11×F – 8½×C = 8½×C + 3×F + 11×F – 8½×C which becomes 14×C = 14×F Divide both sides by 14: 14×C ÷ 14 = 14×F ÷ 14 which makes C = F
Hint #9
Substitute C for F in eq.2a: 2×C – C = A which makes C = A
Hint #10
Substitute C for F in eq.5d: B = 1½×C – C which makes B = ½×C
Hint #11
Substitute C for F in eq.3c: D = ¾×C + ½×C which makes D = 1¼×C
Hint #12
Substitute C for F in eq.4a: E = C – C which makes E = 0
Solution
Substitute C for A and F, ½×C for B, 1¼×C for D, and 0 for E in eq.1: C + ½×C + C + 1¼×C + 0 + C = 19 which simplifies to 4¾×C = 19 Divide both sides of the equation above by 4¾: 4¾×C ÷ 4¾ = 19 ÷ 4¾ which makes C = 4 making A = F = C = 4 B = ½×C = ½ × 4 = 2 D = 1¼×C = 1¼ × 4 = 5 and ABCDEF = 424504