Puzzle for February 1, 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.
* BC and CD are 2-digit numbers (not B×C or C×D).
Scratchpad
Help Area
Hint #1
In eq.3, replace A with E + F (from eq.2): B + C = E + F + E which becomes eq.3a) B + C = 2×E + F Add D to both sides of eq.4: C + D + D = A - D + D which becomes eq.4a) C + 2×D = A
Hint #2
In eq.5, replace A with C + 2×D (from eq.4a): C + 2×D + C = B - C + D + F Add (C - D) to each side of the equation above: C + 2×D + C + C - D = B - C + D + F + C - D which simplifies to 3×C + D = B + F Subtract 3×C from both sides: 3×C + D - 3×C = B + F - 3×C which means eq.5a) D = B + F - 3×C
Hint #3
Subtract F from both sides of eq.2: A - F = E + F - F which becomes A - F = E In eq.3a, substitute (A - F) for E: B + C = 2×(A - F) + F which is the same as B + C = 2×A - 2×F + F which becomes B + C = 2×A - F In the above equation, substitute (C + 2×D) for A (from eq.4a): B + C = 2×(C + 2×D) - F which is the same as B + C = 2×C + 4×D - F Add (F - B - C) to each side: B + C + (F - B - C) = 2×C + 4×D - F + (F - B - C) which simplifies to eq.3b) F = C + 4×D - B
Hint #4
In eq.5a, substitute C + 4×D - B for F (from eq.3b): D = B + C + 4×D - B - 3×C which becomes D = -2×C + 4×D Add (2×C - D) to each side of the above equation: D + 2×C - D = -2×C + 4×D + 2×C - D which simplifies to 2×C = 3×D Divide both sides by 2: 2×C ÷ 2 = 3×D ÷ 2 which means C = 1½×D
Hint #5
Substitute 1½×D for C in eq.4a: 1½×D + 2×D = A which means 3½×D = A
Hint #6
eq.6 can be written as: 10×B + C - (10×C + D) = A + B which is the same as 10×B + C - 10×C - D = A + B Subtract B from both sides of the above equation: 10×B + C - 10×C - D - B = A + B - B which becomes 9×B - 9×C - D = A Substitute 3½×D for A, and (1½×D) for C: 9×B - 9×(1½×D) - D = 3½×D which becomes 9×B - 13½×D - D = 3½×D Add 14½×D to both sides: 9×B - 13½×D - D + 14½×D = 3½×D + 14½×D which simplifies to 9×B = 18×D Divide both sides by 9: 9×B ÷ 9 = 18×D ÷ 9 which makes B = 2×D
Hint #7
Substitute 2×D for B, and (1½×D) for C in eq.5a: D = 2×D + F - 3×(1½×D) which becomes D = 2×D + F - 4½×D which becomes D = F - 2½×D Add 2½×D to both sides: D + 2½×D = F - 2½×D + 2½×D which makes 3½×D = F
Hint #8
Substitute 3½×D for both A and F in eq.2: 3½×D = E + 3½×D Subtract 3½×D from each side: 3½×D - 3½×D = E + 3½×D - 3½×D which means 0 = E
Solution
Substitute 3½×D for A and F, 2×D for B, 1½×D for C, and 0 for E in eq.1: 3½×D + 2×D + 1½×D + D + 0 + 3½×D = 23 which simplifies to 11½×D = 23 Divide both sides by 11½: 11½×D ÷ 11½ = 23 ÷ 11½ which means D = 2 making A = F = 3½×D = 3½ × 2 = 7 B = 2×D = 2 × 2 = 4 C = 1½×D = 1½ × 2 = 3 and ABCDEF = 743207