Puzzle for June 16, 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.
* DE is a 2-digit number (not D×E).
Scratchpad
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Hint #1
eq.4 may be written as: B + E + C = D + F In the above equation, replace B + E with C + D (from eq.2): C + D + C = D + F Subtract D from both sides of the above equation: C + D + C - D = D + F - D which simplifies to eq.4a) 2×C = F
Hint #2
In eq.3, replace F with 2×C: E + 2×C = A + C Subtract C from each side of the above equation: E + 2×C - C = A + C - C which becomes eq.3a) E + C = A
Hint #3
eq.5 may be written as: 10×D + E = A - B + C - D + F In the above equation, replace A with E + C (from eq.3a), and replace F with 2×C: 10×D + E = E + C - B + C - D + 2×C which becomes 10×D + E = E - B - D + 4×C Subtract E from both sides, and add D to both sides: 10×D + E - E + D = E - B - D + 4×C - E + D which simplifies to 11×D = -B + 4×C To express B as a function of C and D, add B to each side, and subtract 11×D from each side: 11×D + B - 11×D = -B + 4×C + B - 11×D which becomes eq.5a) B = 4×C - 11×D
Hint #4
In eq.2, substitute 4×C - 11×D for B (from eq.5a): C + D = 4×C - 11×D + E To express E as a function of C and D, add 11×D to each side, and subtract 4×C from each side: C + D + 11×D - 4×C = 4×C - 11×D + E + 11×D - 4×C which simplifies to eq.2a) 12×D - 3×C = E
Hint #5
To express A as a function of C and D, substitute 12×D - 3×C for E (from eq.2a) in eq.3a: 12×D - 3×C + C = A which becomes eq.3b) 12×D - 2×C = A
Hint #6
Substitute 12×D - 2×C for A (from eq.3b), 4×C - 11×D for B (from eq.5a), 12×D - 3×C for E (from eq.2a), and 2×C for F in eq.1: 12×D - 2×C + 4×C - 11×D + C + D + 12×D - 3×C + 2×C = 22 which becomes 14×D + 2×C = 22 Divide both sides of the above equation by 2: (14×D + 2×C) ÷ 2 = 22 ÷ 2 which becomes 7×D + C = 11 To express C as a function of D, subtract 7×D from both sides: 7×D + C - 7×D = 11 - 7×D which makes eq.1a) C = 11 - 7×D
Hint #7
Substitute (11 - 7×D) for C (from eq.1a) in eq.2a: 12×D - 3×(11 - 7×D) = E which becomes 12×D - 33 + 21×D = E which makes eq.2b) 33×D - 33 = E
Hint #8
To make eq.2b true, check several possible values for D and E: If D = 0, then E = 33×0 - 33 = 0 - 33 = -33 If D = 1, then E = 33×1 - 33 = 33 - 33 = 0 If D = 2, then E = 33×2 - 33 = 66 - 33 = 33 If D > 2, then E > 33 Since E must be a one-digit non-negative integer, then E = 0 which means D = 1
Solution
Substitute 1 for D in eq.1a: C = 11 - 7×1 which becomes C = 11 - 7 which means C = 4 making F = 2×C = 2×4 = 8 A = 12×D - 2×C = 12×1 - 2×4 = 12 - 8 = 4 (from eq.3b) B = 4×C - 11×D = 4×4 - 11×1 = 16 - 11 = 5 (from eq.5a) and ABCDEF = 454108