Puzzle for March 3, 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 EF are 2-digit numbers (not B×C or E×F).
Scratchpad
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Hint #1
Add E to both sides of eq.6: B - E + F + E = C + E + E which becomes eq.6a) B + F = C + 2×E Add the left and right sides of eq.3 to the left and right sides of eq.6a, respectively: B + F + (F - B) = C + 2×E + (A - E) which becomes eq.6b) 2×F = C + E + A
Hint #2
Subtract both B and 2×E from each side of eq.6a: B + F - B - 2×E = C + 2×E - B - 2×E which becomes F - 2×E = C - B In the above equation, replace C - B with B + D + E (from eq.4): F - 2×E = B + D + E Subtract E from both sides: F - 2×E - E = B + D + E - E which becomes eq.6c) F - 3×E = B + D
Hint #3
eq.1 may be re-written as: C + E + A + B + D + F = 30 In the above equation, replace C + E + A with 2×F (from eq.6b): 2×F + B + D + F = 30 which becomes B + D + 3×F = 30 Replace B + D with F - 3×E (from eq.6c): F - 3×E + 3×F = 30 which becomes eq.1a) 4×F - 3×E = 30
Hint #4
Add B + E to both sides of eq.3: F - B + B + E = A - E + B + E which becomes F + E = A + B Substitute F + E for A + B, and A for C + D (from eq.2) in eq.1: F + E + A + E + F = 30 which becomes eq.1b) A + 2×E + 2×F = 30
Hint #5
In eq.1b, substitute 4×F - 3×E for 30 (from eq.1a): A + 2×E + 2×F = 4×F - 3×E Subtract both 2×E and 2×F from each side of the above equation: A + 2×E + 2×F - 2×E - 2×F = 4×F - 3×E - 2×E - 2×F which simplifies to eq.1c) A = 2×F - 5×E
Hint #6
Substitute 2×F - 5×E for A (from eq.1c) in eq.6b: 2×F = C + E + 2×F - 5×E which becomes 2×F = C + 2×F - 4×E Add (4×E - 2×F) to both sides: 2×F + (4×E - 2×F) = C + 2×F - 4×E + (4×E - 2×F) which makes 4×E = C
Hint #7
Substitute 2×F - 5×E for A (from eq.1c) in eq.3: F - B = 2×F - 5×E - E which becomes F - B = 2×F - 6×E Subtract F from both sides of the above equation: F - B - F = 2×F - 6×E - F which becomes -B = F - 6×E Multiply both sides by (-1): -B × (-1) = (F - 6×E) × (-1) which becomes B = -F + 6×E which is equivalent to eq.3a) B = 6×E - F
Hint #8
Substitute 2×F - 5×E for A (from eq.1c), and 4×E for C in eq.2: 4×E + D = 2×F - 5×E Subtract 4×E from each side of the above equation: 4×E + D - 4×E = 2×F - 5×E - 4×E which becomes eq.2a) D = 2×F - 9×E
Hint #9
eq.5 can be written as: 10×B + C - F = D + 10×E + F Add F to both sides: 10×B + C - F + F = D + 10×E + F + F which becomes 10×B + C = D + 10×E + 2×F Substitute (6×E - F) for B (from eq.3a), 4×E for C, and 2×F - 9×E for D (from eq.2a) in the above equation: 10×(6×E - F) + 4×E = 2×F - 9×E + 10×E + 2×F which becomes 60×E - 10×F + 4×E = 4×F + E Add (10×F - E) to each side: 60×E - 10×F + 4×E + (10×F - E) = 4×F + E + (10×F - E) which simplifies to 63×E = 14×F Divide both sides by 14: 63×E ÷ 14 = 14×F ÷ 14 which means 4½×E = F
Hint #10
Substitute 4½×E for F in eq.1c: A = 2×(4½×E) - 5×E which becomes A = 9×E - 5×E which makes A = 4×E
Hint #11
Substitute 4½×E for F in eq.3a: B = 6×E - 4½×E which makes B = 1½×E
Hint #12
Substitute (4½×E) for F in eq.2a: D = 2×(4½×E) - 9×E which becomes D = 9×E - 9×E which makes D = 0
Solution
Substitute 4×E for A and C, 1½×E for B, 0 for D, and 4½×E for F in eq.1: 4×E + 1½×E + 4×E + 0 + E + 4½×E = 30 which simplifies to 15×E = 30 Divide both sides by 15: 15×E ÷ 15 = 30 ÷ 15 which means E = 2 making A = C = 4×E = 4 × 2 = 8 B = 1½×E = 1½ × 2 = 3 F = 4½×E = 4½ × 2 = 9 and ABCDEF = 838029