Puzzle for March 17, 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 DE are 2-digit numbers (not B×C or D×E). DEF is a 3-digit number (not D×E×F).
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Hint #1
Add B + D to both sides of eq.2: C - B + B + D = E - D + B + D which becomes C + D = E + B which may be written as eq.2a) E + B = C + D
Hint #2
Subtract the left and right sides of eq.2a from the left and right sides of eq.4, respectively: B - A - (E + B) = C - D - (C + D) which is the same as B - A - E - B = C - D - C - D which becomes -A - E = -2×D Add E to both sides of the equation above: -A - E + E = -2×D + E which becomes -A = -2×D + E Multiply both sides by (-1): -A × (-1) = (-2×D + E) × (-1) which makes eq.4a) A = 2×D - E
Hint #3
eq.4 may be written as: C - D = B - A Subtract the left and right sides of the above equation from the left and right sides of eq.3, respectively: C - A - (C - D) = B - F - (B - A) which is the same as C - A - C + D = B - F - B + A which becomes -A + D = -F + A Add A + F to both sides of the equation above: -A + D + A + F = -F + A + A + F which becomes eq.4b) D + F = 2×A
Hint #4
In eq.4b, substitute (2×D - E) for A (from eq.4a): D + F = 2×(2×D - E) which is equivalent to D + F = 4×D - 2×E Subtract D from each side: D + F - D = 4×D - 2×E - D which makes eq.3a) F = 3×D - 2×E
Hint #5
eq.5 may be written as: 10×D + E = B + F - D In the above equation, substitute (3×D - 2×E) for F (from eq.3a): 10×D + E = B + (3×D - 2×E) - D which becomes 10×D + E = B + 2×D - 2×E Add (2×E - 2×D) to each side: 10×D + E + (2×E - 2×D) = B + 2×D - 2×E + (2×E - 2×D) which simplifies to eq.5a) 8×D + 3×E = B
Hint #6
Substitute (2×D - E) for A (from eq.4a), 8×D + 3×E for B (from eq.5a), and (3×D - 2×E) for F (from eq.3a) into eq.3: C - (2×D - E) = 8×D + 3×E - (3×D - 2×E) which becomes C - 2×D + E = 5×D + 5×E Add (2×D - E) to each side of the equation above: C - 2×D + E + (2×D - E) = 5×D + 5×E + (2×D - E) which becomes eq.3b) C = 7×D + 4×E
Hint #7
eq.6 may be written as: 10×B + C = 100×D + 10×E + F - B - C - D Add B + C to both sides of the above equation: 10×B + C + B + C = 100×D + 10×E + F - B - C - D + B + C which becomes eq.6a) 11×B + 2×C = 99×D + 10×E + F
Hint #8
Substitute (8×D + 3×E) for B (from eq.5a), (7×D + 4×E) for C (from eq.3b), and (3×D - 2×E) for F (from eq.3a) into eq.6a: 11×(8×D + 3×E) + 2×(7×D + 4×E) = 99×D + 10×E + (3×D - 2×E) which becomes 88×D + 33×E + 14×D + 8×E = 102×D + 8×E which becomes 102×D + 41×E = 102×D + 8×E Subtract both 102×D and 8×E from each side: 102×D + 41×E - 102×D - 8×E = 102×D + 8×E - 102×D - 8×E which simplifies to 33×E = 0 which means E = 0
Hint #9
Substitute 0 for E in eq.4a: A = 2×D - 0 which means A = 2×D
Hint #10
Substitute 0 for E in eq.5a: 8×D + 3×0 = B which means 8×D = B
Hint #11
Substitute 0 for E in eq.3b: C = 7×D + 4×0 which makes C = 7×D
Hint #12
Substitute 0 for E in eq.3a: F = 3×D - 2×0 which means F = 3×D
Solution
Substitute 2×D for A, 8×D for B, 7×D for C, 0 for E, and 3×D for F in eq.1: 2×D + 8×D + 7×D + D + 0 + 3×D = 21 which simplifies to 21×D = 21 Divide both sides by 21: 21×D ÷ 21 = 21 ÷ 21 which means D = 1 making A = 2×D = 2 × 1 = 2 B = 8×D = 8 × 1 = 8 C = 7×D = 7 × 1 = 7 F = 3×D = 3 × 1 = 3 and ABCDEF = 287103