Puzzle for October 27, 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 and EF are 2-digit numbers (not D×E or E×F). BCD is a 3-digit number (not B×C×D).
Scratchpad
Help Area
Hint #1
eq.6 may be written as: 100×B + 10×C + D + C + D + E + F = 10×D + E + 10×E + F which becomes 100×B + 11×C + 2×D + E + F = 10×D + 11×E + F Subtract 2×D, E, and F from each side of the above equation: 100×B + 11×C + 2×D + E + F – 2×D – E – F = 10×D + 11×E + F – 2×D – E – F which simplifies to eq.6a) 100×B + 11×C = 8×D + 10×E
Hint #2
In eq.2, replace F with B + C (from eq.4): E + B + C = A + B Subtract B from both sides of the above equation: E + B + C – B = A + B – B which becomes eq.2a) E + C = A
Hint #3
In eq.3, replace A with E + C (from eq.2a): D = E + C + C which becomes eq.3a) D = E + 2×C
Hint #4
Substitute E + C for A (from eq.2a), E + 2×C for D (from eq.3a), and B + C for F (from eq.4) in eq.5: E + C + E = E + 2×C + B + C which becomes E + C + E = E + 3×C + B Subtract 3×C and E from both sides of the equation above: E + C + E – 3×C – E = E + 3×C + B – 3×C – E which becomes eq.5a) E – 2×C = B
Hint #5
Substitute (E – 2×C) for B (from eq.5a), and (E + 2×C) for D (from eq.3a) in eq.6a: 100×(E – 2×C) + 11×C = 8×(E + 2×C) + 10×E which is the same as 100×E – 200×C + 11×C = 8×E + 16×C + 10×E which becomes 100×E – 189×C = 16×C + 18×E In the equation above, add 189×C to each side, and subtract 18×E from each side: 100×E – 189×C + 189×C – 18×E = 16×C + 18×E + 189×C – 18×E which simplifies to 82×E = 205×C Divide both sides by 82: 82×E ÷ 82 = 205×C ÷ 82 which makes E = 2½×C
Hint #6
Substitute 2½×C for E in eq.2a: 2½×C + C = A which means 3½×C = A
Hint #7
Substitute 2½×C for E in eq.5a: 2½×C – 2×C = B which means ½×C = B
Hint #8
Substitute 2½×C for E in eq.3a: D = 2½×C + 2×C which makes D = 4½×C
Hint #9
Substitute ½×C for B in eq.4: F = ½×C + C which makes F = 1½×C
Solution
Substitute 3½×C for A, ½×C for B, 4½×C for D, 2½×C for E, and 1½×C for F in eq.1: 3½×C + ½×C + C + 4½×C + 2½×C + 1½×C = 27 which becomes 13½×C = 27 Divide both sides of the equation above by 13½: 13½×C ÷ 13½ = 27 ÷ 13½ which means C = 2 making A = 3½×C = 3½ × 2 = 7 B = ½×C = ½ × 2 = 1 D = 4½×C = 4½ × 2 = 9 E = 2½×C = 2½ × 2 = 5 F = 1½×C = 1½ × 2 = 3 and ABCDEF = 712953