Puzzle for January 20, 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 is a 2-digit number (not C×D).
** CDE and DEF are 3-digit numbers (not C×D×E or D×E×F).
Scratchpad
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Hint #1
Add F to both sides of eq.5: CD - F + F = D + E + F + F which becomes CD = D + E + 2×F In eq.4, replace CD with D + E + 2×F: D + E + 2×F = A + B + C + D Replace D + E with B + C (from eq.2) in the above equation: B + C + 2×F = A + B + C + D Subtract B, C, and D from both sides: B + C + 2×F - B - C - D = A + B + C + D - B - C - D which simplifies to eq.5a) 2×F - D = A
Hint #2
Add B, C, D, and E to both sides of eq.3: A - B - C - D - E + B + C + D + E = B + C + D + E + F + B + C + D + E which becomes A = 2×B + 2×C + 2×D + 2×E + F In the above equation, substitute 2×F - D for A (from eq.5a): 2×F - D = 2×B + 2×C + 2×D + 2×E + F Add (D - F) to both sides: 2×F - D + (D - F) = 2×B + 2×C + 2×D + 2×E + F + (D - F) which becomes eq.3a) F = 2×B + 2×C + 3×D + 2×E
Hint #3
eq.3a may be written as: F = 2×(B + C) + 3×D + 2×E In the equation above, substitute D + E for B + C (from eq.2): F = 2×(D + E) + 3×D + 2×E which is equivalent to F = 2×D + 2×E + 3×D + 2×E which means eq.3b) F - 5×D = 4×E
Hint #4
eq.5 may be written as: 10×C + D - F = D + E + F Add (F - D) to each side of the above equation: 10×C + D - F + (F - D) = D + E + F + (F - D) which becomes 10×C = E + 2×F Multiply both sides of the equation above by 4: 4×10×C = 4×(E + 2×F) which becomes 40×C = 4×E + 8×F Replace 4×E with F - 5×D (from eq.3b): 40×C = F - 5D + 8×F which becomes eq.5b) 40×C = 9×F - 5×D
Hint #5
eq.6 may be written as: 100×C + 10×D + E - F = 100×D + 10×E + F Add (F - E - 10×D) to both sides of the above equation: 100×C + 10×D + E - F + (F - E - 10×D) = 100×D + 10×E + F + (F - E - 10×D) which simplifies to 100×C = 90×D + 9×E + 2×F Multiply both sides of this equation by 4: 4×100×C = 4×(90×D + 9×E + 2×F) which is equivalent to 10×(40×C) = 360×D + 9×(4×E) + 8×F
Hint #6
Substitute (9×F - 5×D) for 40×C (from eq.5b), and (F - 5×D) for 4×E (from eq.3b) in the above equation: 10×(9×F - 5×D) = 360×D + 9×(F - 5×D) + 8×F which is the same as 90×F - 50×D = 360×D + 9×F - 45×D + 8×F which becomes 90×F - 50×D = 315×D + 17×F Add (50×D - 17×F) to both sides of the equation above: 90×F - 50×D + (50×D - 17×F) = 315×D + 17×F + (50×D - 17×F) which becomes 73×F = 365×D Divide both sides by 73: 73×F ÷ 73 = 365×D ÷ 73 which makes F = 5×D
Hint #7
In eq.5b, substitute (5×D) for F: 40×C = 9×(5×D) - 5×D which becomes 40×C = 45×D - 5×D = 40×D Divide both sides by 40: 40×C ÷ 40 = 40×D ÷ 40 which means C = D
Hint #8
Substitute 5×D for F in eq.3b: 5×D - 5×D = 4×E which means 0 = 4×E which means 0 = E
Hint #9
Substitute 5×D for F in eq.5a: 2×(5×D)- D = A which means 10×D - D = A which makes 9×D = A
Hint #10
Substitute 0 for E, and D for C in eq.2: B + D = D + 0 Subtract D from each side: B + D - D = D + 0 - D which makes B = 0
Solution
Substitute 9×D for A, 0 for B and E, D for C, and 5×D for F in eq.1: 9×D + 0 + D + D + 0 + 5×D = 16 which simplifies to 16×D = 16 Divide both sides by 16: 16×D ÷ 16 = 16 ÷ 16 which becomes D = 1 making A = 9×D = 9 × 1 = 9 C = D = 1 F = 5×D = 5 × 1 = 5 and ABCDEF = 901105