Puzzle for February 19, 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.
* DE is a 2-digit number (not D×E).
Scratchpad
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Hint #1
eq.5 may be written as: E = (A + C + F) ÷ 3 Multiply both sides of the above equation by 3: 3 × E = 3 × (A + C + F) ÷ 3 which becomes eq.5a) 3×E = A + C + F
Hint #2
eq.1 may be re-written as: A + C + F + B + D + E = 30 In the equation above, replace A + C + F with 3×E (from eq.5a), and B + D with E (from eq.2): 3×E + E + E = 30 which makes 5×E = 30 Divide both sides by 5: 5×E ÷ 5 = 30 ÷ 5 which makes E = 6
Hint #3
In eq.2, replace E with 6: B + D = 6 Subtract D from both sides of the equation above: B + D – D = 6 – D which makes eq.2a) B = 6 – D
Hint #4
eq.3 may be written as: 10×D + E = A + B + C In the equation above, substitute 6 for E, and 6 – D for B (from eq.2a): 10×D + 6 = A + 6 – D + C Subtract 6 from both sides, and add D to both sides: 10×D + 6 – 6 + D = A + 6 – D + C – 6 + D which becomes eq.3a) 11×D = A + C
Hint #5
Substitute 6 for E, and 11×D for A + C (from eq.3a) in eq.5a: 3×6 = 11×D + F which becomes 18 = 11×D + F Subtract 11×D from both sides of the above equation: 18 – 11×D = 11×D + F – 11×D which becomes eq.5b) 18 – 11×D = F
Hint #6
To make eq.5b true, check several possible values for D and F: If D = 0, then F = 18 – 11×0 = 18 – 0 = 18 If D = 1, then F = 18 – 11×1 = 18 – 11 = 7 If D = 2, then F = 18 – 11×2 = 18 – 22 = –4 If D > 2, then F < –4 Since F must be a non-negative one-digit integer, then the only values for D and F that make eq.5b true are: D = 1 and F = 7 making B = 6 – D = 6 – 1 = 5 (from eq.2a)
Hint #7
Substitute 1 for D in eq.3a: 11×1 = A + C which becomes 11 = A + C Subtract A from each side of the equation above: 11 – A = A + C – A which becomes eq.3b) 11 – A = C
Hint #8
Substitute 6 for E, and 5 for B in eq.4: C ÷ 6 = A ÷ 5 Multiply each side of the above equation by both 6 and 5: (C ÷ 6) × 6 × 5 = (A ÷ 5) × 6 × 5 which becomes C × 5 = A × 6 which may be written as eq.4a) 5×C = 6×A
Solution
Substitute (11 – A) for C (from eq.3b) in eq.4a: 5×(11 – A) = 6×A which becomes 5×11 – 5×A = 6×A which becomes 55 – 5×A = 6×A Add 5×A to both sides of the equation above: 55 – 5×A + 5×A = 6×A + 5×A which becomes 55 = 11×A Divide both sides by 11: 55 ÷ 11 = 11×A ÷ 11 which means 5 = A making C = 11 – A = 11 – 5 = 6 (from eq.3b) and ABCDEF = 556167