Puzzle for August 11, 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.
* AB and CD are 2-digit numbers (not A×B or C×D).
Scratchpad
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Hint #1
Add E to both sides of eq.4: D + C – E + E = B + F + E which becomes D + C = B + F + E which may be written as D + C = B + E + F In the above equation, replace B + E with C + F (from eq.2): D + C = C + F + F Subtract C from both sides: D + C – C = C + F + F – C which means eq.4a) D = 2×F
Hint #2
In eq.3, replace D with 2×F: B + 2×F = A + F Subtract F from both sides of the above equation: B + 2×F – F = A + F – F which becomes eq.3a) B + F = A
Hint #3
In eq.5, substitute A for B + F (from eq.3a): C + E = A + A which becomes eq.5a) C + E = 2×A
Hint #4
eq.1 may be written as: A + B + F + D + C + E = 28 Substitute A for B + F (from eq.3a), and 2×A for C + E (from eq.5a) in the above equation: A + A + D + 2×A = 28 which becomes 4×A + D = 28 Subtract D from both sides of the above equation: 4×A + D – D = 28 – D which becomes 4×A = 28 – D Divide both sides by 4: 4×A ÷ 4 = (28 – D) ÷ 4 which means eq.1a) A = 7 – ¼×D
Hint #5
Divide both sides of eq.4a by 2: D ÷ 2 = 2×F ÷ 2 which makes ½×D = F Substitute 7 – ¼×D for A (from eq.1a), and ½×D for F in eq.3: B + D = 7 – ¼×D + ½×D Subtract D from both sides of the above equation: B + D – D = 7 – ¼×D + ½×D – D which makes eq.4b) B = 7 – ¾×D
Hint #6
Subtract the left and right sides of eq.4 from the left and right sides of eq.5, respectively: C + E – (D + C – E) = A + B + F – (B + F) which is equivalent to C + E – D – C + E = A + B + F – B – F which becomes 2×E – D = A Substitute 7 – ¼×D for A (from eq.1a) in the above equation: 2×E – D = 7 – ¼×D Add D to each side: 2×E – D + D = 7 – ¼×D + D which becomes 2×E = 7 + ¾×D Divide both sides by 2: 2×E ÷ 2 = (7 + ¾×D) ÷ 2 which means eq.5b) E = 3½ + ⅜×D
Hint #7
Substitute ½×D for F, 7 – ¾×D for B (from eq.4b), and 3½ + ⅜×D for E (from eq.5b) in eq.2: C + ½×D = 7 – ¾×D + 3½ + ⅜×D which becomes C + ½×D = 10½ – ⅜×D Subtract ½×D from both sides: C + ½×D – ½×D = 10½ – ⅜×D – ½×D which means eq.2a) C = 10½ – ⅞×D
Hint #8
eq.6 may be written as: 10×A + B + E = 10×C + D – E Add E to both sides of the above equation: 10×A + B + E + E = 10×C + D – E + E which becomes eq.6a) 10×A + B + 2×E = 10×C + D
Solution
Substitute (7 – ¼×D) for A (from eq.1a), 7 – ¾×D for B (from eq.4b), (3½ + ⅜×D) for E (from eq.5b), and (10½ – ⅞×D) for C (from eq.2a) in eq.6a: 10×(7 – ¼×D) + 7 – ¾×D + 2×(3½ + ⅜×D) = 10×(10½ – ⅞×D) + D which becomes 70 – 2½×D + 7 – ¾×D + 7 + ¾×D = 105 – 8¾×D + D which becomes 84 – 2½×D = 105 – 7¾×D Add 7¾×D to both sides, and subtract 84 from each side: 84 – 2½×D + 7¾×D – 84 = 105 – 7¾×D + 7¾×D – 84 which makes 5¼×D = 21 Divide both sides by 5¼: 5¼×D ÷ 5¼ = 21 ÷ 5¼ which means D = 4 making A = 7 – ¼×D = 7 – ¼ × 4 = 7 – 1 = 6 (from eq.1a) B = 7 – ¾×D = 7 – ¾ × 4 = 7 – 3 = 4 (from eq.4b) C = 10½ – ⅞×D = 10½ – ⅞ × 4 = 10½ – 3½ = 7 (from eq.2a) E = 3½ + ⅜×D = 3½ + ⅜ × 4 = 3½ + 1½ = 5 (from eq.5b) F = ½×D = ½ × 4 = 2 and ABCDEF = 647452