Puzzle for July 14, 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.
* CD, DE, and EF are 2-digit numbers (not C×D, D×E, or E×F).
Scratchpad
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Hint #1
Add D to both sides of eq.4: D + E + D = B - C - D + D which becomes 2×D + E = B - C Subtract the left and right sides of the above equation from the left and right sides of eq.2, respectively: E + F - (2×D + E) = B + C + D - (B - C) which is the same as E + F - 2×D - E = B + C + D - B + C which becomes F - 2×D = 2×C + D Subtract D from both sides: F - 2×D - D = 2×C + D - D which becomes eq.2a) F - 3×D = 2×C
Hint #2
Add C to both sides of eq.5: F - C + C = A + C + E + C which becomes F = A + 2×C + E In the above equation, replace 2×C with F - 3×D (from eq.2a): F = A + F - 3×D + E Add 3×D to both sides, and subtract F from both sides: F + 3×D - F = A + F - 3×D + E + 3×D - F which becomes eq.5a) 3×D = A + E
Hint #3
In eq.3, replace A + E with B - D (from eq.5a): B - D = 3×D Add D to both sides: B - D + D = 3×D + D which becomes eq.5b) B = 4×D
Hint #4
eq.4 may be written as: D + E = B - D - C In the above equation, substitute A + E for B - D (from eq.3): D + E = A + E - C Subtract E from both sides: D + E - E = A + E - C - E D = A - C Express A as a function of C and D by adding C to each side: D + C = A - C + C which becomes eq.4a) D + C = A
Hint #5
Substitute 4×D for B (from eq.5b) in eq.4: D + E = 4×D - C - D which becomes D + E = 3×D - C Express E as a function of C and D by subtracting D from each side: D + E - D = 3×D - C - D which becomes eq.4b) E = 2×D - C
Hint #6
Express F as a function of C and D by adding 3×D to each side of eq.2a: F - 3×D + 3×D = 2×C + 3×D which becomes eq.2b) F = 2×C + 3×D
Hint #7
eq.6 may be written as: A + 10×D + E = 10×E + F - (10×C + D) which may be written as A + 10×D + E = 10×E + F - 10×C - D In the above equation, subtract E from both sides, and add D to each side: A + 10×D + E - E + D = 10×E + F - 10×C - D - E + D which becomes eq.6a) A + 11×D = 9×E + F - 10×C
Hint #8
Substitute D + C for A (from eq.4a), (2×D - C) for E (from eq.4b), and 2×C + 3×D for F (from eq.2b) in eq.6a: D + C + 11×D = 9×(2×D - C) + 2×C + 3×D - 10×C which is equivalent to D + C + 11×D = 18×D - 9×C + 2×C + 3×D - 10×C which becomes C + 12×D = 21×D - 17×C Subtract 12×D from both sides, and add 17×C to each side: C + 12×D - 12×D + 17×C = 21×D - 17×C - 12×D + 17×C which simplifies to 18×C = 9×D Divide both sides by 9: 18×C ÷ 9 = 9×D ÷ 9 which makes 2×C = D
Hint #9
Substitute 2×C for D (in eq.4a): 2×C + C = A which makes 3×C = A
Hint #10
Substitute 2×C for D (in eq.5b): B = 4×(2×C) which makes B = 8×C
Hint #11
Substitute (2×C) for D (in eq.4b): E = 2×(2×C) - C which becomes E = 4×C - C which makes E = 3×C
Hint #12
Substitute (2×C) for D (in eq.2b): F = 2×C + 3×(2×C) which is equivalent to F = 2×C + 6×C which makes F = 8×C
Solution
Substitute 3×C for A and E, 8×C for B and F, and 2×C for D in eq.1: 3×C + 8×C + C + 2×C + 3×C + 8×C = 25 which simplifies to 25×C = 25 Divide both sides by 25: 25×C ÷ 25 = 25 ÷ 25 which means C = 1 making A = E = 3×C = 3×1 = 3 B = F = 8×C = 8×1 = 8 D = 2×C = 2×1 = 2 and ABCDEF = 381238