Puzzle for December 2, 2018 ( )
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.
Scratchpad
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Hint #1
eq.5 may be written as: A - E = D - F Add the left and right sides of the above equation to the respective sides of eq.4: A + E + (A - E) = B + F + (D - F) which becomes eq.4a) 2×A = B + D Subtract B from each side of the above equation: 2×A - B = B + D - B which becomes eq.4b) 2×A - B = D Add E to each side of eq.6: A + B + E + E = C - E + E which becomes eq.6a) A + B + 2×E = C
Hint #2
Substitute A + B + 2×E for C (from eq.6a) in eq.3: A + D = A + B + 2×E + E Subtract A from both sides of the equation above: A + D - A = A + B + 2×E + E - A which simplifies to eq.3a) D = B + 3×E
Hint #3
Substitute B + 3×E for D (from eq.3a) in eq.4a: 2×A = B + B + 3×E Subtract 3×E from each side: 2×A - 3×E = B + B + 3×E - 3×E which becomes eq.4c) 2×A - 3×E = 2×B
Hint #4
Substitute 2×A for B + D (from eq.4a), and B + C for E + F (from eq.2) in eq.1: A + 2×A + C + B + C = 23 which becomes 3×A + B + 2×C = 23 From eq.6a, substitute (A + B + 2×E) for C in the above equation: 3×A + B + 2×(A + B + 2×E) = 23 which is the same as 3×A + B + 2×A + 2×B + 4×E = 23 which becomes eq.1a) 5×A + 3×B + 4×E = 23
Hint #5
Substitute A + E for B + F (from eq.4), and A + D for C + E (from eq.3) in eq.1: A + A + E + A + D + D = 23 which becomes 3×A + 2×D + E = 23 From eq.4b, substitute (2×A - B) for D in the above equation: 3×A + 2×(2×A - B) + E = 23 which is the same as 3×A + 4×A - 2×B + E = 23 which becomes eq.1b) 7×A - 2×B + E = 23
Hint #6
From eq.4c, substitute (2×A - 3×E) for 2×B in eq.1b: 7×A - (2×A - 3×E) + E = 23 which is the same as 7×A - 2×A + 3×E + E = 23 which becomes 5×A + 4×E = 23 Substitute 5×A + 3×B + 4×E for 23 (from eq.1a) in the above equation: 5×A + 4×E = 5×A + 3×B + 4×E Subtract both 5×A and 4×E from each side: 5×A + 4×E - 5×A - 4×E = 5×A + 3×B + 4×E - 5×A - 4×E which simplifies to 0 = 3×B which means 0 = B
Hint #7
Replace B with 0 in eq.3a: D = 0 + 3×E which means D = 3×E
Hint #8
Substitute 0 for B, and 3×E for D in eq.4a: 2×A = 0 + 3×E which makes 2×A = 3×E Divide both sides by 2: 2×A ÷ 2 = 3×E ÷ 2 which makes A = 1½×E
Hint #9
Substitute 1½×E for A, and 0 for B in eq.6a: 1½×E + 0 + 2×E = C which makes 3½×E = C
Hint #10
Substitute 0 for B, and 3½×E for C in eq.2: 0 + 3½×E = E + F Subtract E from both sides: 0 + 3½×E - E = E + F - E which makes 2½×E = F
Solution
Substitute 1½×E for A, 0 for B, 3½×E for C, 3×E for D, and 2½×E for F in eq.1: 1½×E + 0 + 3½×E + 3×E + E + 2½×E = 23 which simplifies to 11½×E = 23 Divide both sides by 11½: 11½×E ÷ 11½ = 23 ÷ 11½ which means E = 2 making A = 1½×E = 1½ × 2 = 3 C = 3½×E = 3½ × 2 = 7 D = 3×E = 3 × 2 = 6 F = 2½×E = 2½ × 2 = 5 and ABCDEF = 307625