Puzzle for October 16, 2022 ( )
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.
* AB, DE, and EF are 2-digit numbers (not A×B, D×E, or E×F).
Scratchpad
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Hint #1
Add F and E to both sides of eq.3: E - F + F + E = C - E + F + E which becomes eq.3a) 2×E = C + F
Hint #2
eq.6 may be written as: E = (A + C + D + F) ÷ 4 Multiply both sides of the above equation by 4: 4 × E = 4 × (A + C + D + F) ÷ 4 which becomes 4×E = A + C + D + F which is the same as eq.6a) 4×E = A + D + C + F
Hint #3
In eq.6a, replace C + F with 2×E (from eq.3a): 4×E = A + D + 2×E Subtract 2×E from each side of the above equation: 4×E - 2×E = A + D + 2×E - 2×E which becomes eq.6b) 2×E = A + D
Hint #4
In eq.6b, substitute C + F for 2×E (from eq.3a), and B + F for A + D (from eq.2): C + F = B + F Subtract F from each side of the equation above: C + F - F = B + F - F which makes C = B
Hint #5
eq.1 may be written as: A + D + C + F + E + B = 20 In the above equation, substitute 4×E for A + D + C + F (from eq.6a): 4×E + E + B = 20 which becomes 5×E + B = 20 Subtract 5×E from both sides: 5×E + B - 5×E = 20 - 5×E which makes B = 20 - 5×E and also makes eq.1a) C = B = 20 - 5×E
Hint #6
Substitute 20 - 5×E for C (from eq.1a) in eq.3a: 2×E = 20 - 5×E + F In the above equation, subtract 20 from both sides, and add 5×E to both sides: 2×E - 20 + 5×E = 20 - 5×E + F - 20 + 5×E which becomes eq.3b) 7×E - 20 = F
Hint #7
eq.4 may be written as: 10×A + B - F = 10×D + E + 10×E + F which becomes 10×A + B - F = 10×D + 11×E + F Add F to both sides of the above equation: 10×A + B - F + F = 10×D + 11×E + F + F which becomes eq.4a) 10×A + B = 10×D + 11×E + 2×F
Hint #8
Substitute 20 - 5×E for B (from eq.1a), and (7×E - 20) for F (from eq.3b) in eq.4a: 10×A + 20 - 5×E = 10×D + 11×E + 2×(7×E - 20) which becomes 10×A + 20 - 5×E = 10×D + 11×E + 14×E - 40 which becomes 10×A + 20 - 5×E = 10×D + 25×E - 40 In the above equation, subtract 20 from both sides, and add 5×E to both sides: 10×A + 20 - 5×E - 20 + 5×E = 10×D + 25×E - 40 - 20 + 5×E which becomes 10×A = 10×D + 30×E - 60 Divide both sides by 10: 10×A ÷ 10 = (10×D + 30×E - 60) ÷ 10 which becomes eq.4b) A = D + 3×E - 6
Hint #9
Substitute D + 3×E - 6 for A (from eq.4b) into eq.6b: 2×E = D + 3×E - 6 + D which becomes 2×E = 2×D + 3×E - 6 In the above equation, subtract 3×E from both sides, and add 6 to both sides: 2×E - 3×E + 6 = 2×D + 3×E - 6 - 3×E + 6 which becomes -E + 6 = 2×D which is the same as 6 - E = 2×D Divide both sides by 2: (6 - E) ÷ 2 = 2×D ÷ 2 which becomes eq.6c) 3 - ½×E = D
Hint #10
Substitute 3 - ½×E for D (from eq.6c) in eq.4b: A = 3 - ½×E + 3×E - 6 which becomes A = -3 + 2½×E which is the same as eq.4c) A = 2½×E - 3
Hint #11
eq.5 may be written as: 10×D + E = A - D + F Substitute (3 - ½×E) for D (from eq.6c), 2½×E - 3 for A (from eq.4c), and 7×E - 20 for F (from eq.3b) in the equation above: 10×(3 - ½×E) + E = 2½×E - 3 - (3 - ½×E) + 7×E - 20 which becomes 30 - 5×E + E = 2½×E - 3 - 3 + ½×E + 7×E - 20 which becomes eq.5a) 30 - 4×E = 10×E - 26
Hint #12
Add 4×E and 26 to both sides of eq.5a: 30 - 4×E + 4×E + 26 = 10×E - 26 + 4×E + 26 which makes 56 = 14×E Divide both sides by 14: 56 ÷ 14 = 14×E ÷ 14 which makes 4 = E
Solution
Since E = 4, then: A = 2½×E - 3 = (2½ × 4) - 3 = 10 - 3 = 7 (from eq.4c) C = B = 20 - 5×E = 20 - (5 × 4) = 20 - 20 = 0 (from eq.1a) D = 3 - ½×E = 3 - (½ × 4) = 3 - 2 = 1 (from eq.6c) F = 7×E - 20 = (7 × 4) - 20 = 28 - 20 = 8 (from eq.3b) and ABCDEF = 700148