Puzzle for January 26, 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, DE, and EF are 2-digit numbers (not B×C, D×E, or E×F).
Scratchpad
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Hint #1
In eq.5, replace C with A - B (from eq.3): A - E = A - B - D + F In the above equation, substitute (D + E) for B (from eq.2): A - E = A - (D + E) - D + F which is equivalent to A - E = A - D - E - D + F which becomes A - E = A - E - 2×D + F Subtract (A - E) from each side: A - E - (A - E) = A - E - 2×D + F - (A - E) which becomes 0 = -2×D + F Add 2×D to both sides: 0 + 2×D = -2×D + F + 2×D which means 2×D = F
Hint #2
In eq.4, replace F with 2×D: D + 2×D = A - D Add D to each side of the above equation: D + 2×D + D = A - D + D which makes 4×D = A
Hint #3
Substitute 4×D for A, and 2×D for F in eq.5: 4×D - E = C - D + 2×D which becomes 4×D - E = C + D Add (E - D) to each side of the above equation: 4×D - E + (E - D) = C + D + (E - D) which simplifies to eq.5a) 3×D = C + E
Hint #4
Subtract D from both sides of eq.2: B - D = D + E - D which becomes eq.2a) B - D = E
Hint #5
eq.6 may be written as: 10×B + C - (10×D + E) = 10×E + F which becomes 10×B + C - 10×D - E = 10×E + F Add E to each side of the above equation: 10×B + C - 10×D - E + E = 10×E + F + E which becomes 10×B + C - 10×D = 11×E + F which may be written as eq.6a) 10×(B - D) + C = 11×E + F
Hint #6
In eq.6a, replace (B - D) with E (from eq.2a): 10×E + C = 11×E + F Subtract 10×E from each side: 10×E + C - 10×E = 11×E + F - 10×E which becomes eq.6b) C = E + F
Hint #7
In eq.6b, replace F with 2×D: C = E + 2×D In eq.5a, replace C with E + 2×D: 3×D = E + 2×D + E Subtract 2×D from each side: 3×D - 2×D = E + 2×D + E - 2×D which becomes D = 2×E Divide both sides by 2: D ÷ 2 = 2×E ÷ 2 which makes ½×D = E
Hint #8
Substitute ½×D for E in eq.5a: 3×D = C + ½×D Subtract ½×D from each side: 3×D - ½×D = C + ½×D - ½×D which makes 2½×D = C
Hint #9
Substitute ½×D for E in eq.2: B = D + ½×D which means B = 1½×D
Solution
Substitute 4×D for A, 1½×D for B, 2½×D for C, ½×D for E, and 2×D for F in eq.1: 4×D + 1½×D + 2½×D + D + ½×D + 2×D = 23 which simplifies to 11½×D = 23 Divide both sides by 11½: 11½×D ÷ 11½ = 23 ÷ 11½ which makes D = 2 making A = 4×D = 4 × 2 = 8 B = 1½×D = 1½ × 2 = 3 C = 2½×D = 2½ × 2 = 5 E = ½×D = ½ × 2 = 1 F = 2×D = 2 × 2 = 4 and ABCDEF = 835214