Puzzle for January 15, 2023 ( )
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.2 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 eq.2a) 4×E = A + C + D + F
Hint #2
eq.3 may be written as: F = (A + B + C + E) ÷ 4 Multiply both sides of the above equation by 4: 4 × F = 4 × (A + B + C + E) ÷ 4 which becomes eq.3a) 4×F = A + B + C + E
Hint #3
Subtract the left and right sides of eq.3a from the left and right sides of eq.2a, respectively: 4×E - 4×F = A + C + D + F - (A + B + C + E) which becomes 4×E - 4×F = A + C + D + F - A - B - C - E which becomes 4×E - 4×F = D + F - B - E In the above equation, subtract F from both sides, and add E to both sides: 4×E - 4×F - F + E = D + F - B - E - F + E which becomes eq.2b) 5×E - 5×F = D - B
Hint #4
Add E and F to both sides of eq.1: A - E + E + F = E - F + E + F which becomes eq.1a) A + F = 2×E
Hint #5
Multiply both sides of eq.6 by D and F: (C ÷ D) × (D × F) = (A ÷ F) × (D × F) which becomes eq.6a) C × F = A × D Since both C and D ≠ 0 (from eq.4 and eq.5), divide both sides of eq.6a by (C × D): (C × F) ÷ (C × D) = (A × D) ÷ (C × D) which becomes eq.6b) F ÷ D = A ÷ C
Hint #6
In eq.6b, replace F ÷ D with A - B - E (from eq.5), and A ÷ C with E - D (from eq.4): A - B - E = E - D Add E to both sides of the equation above: A - B - E + E = E - D + E which becomes eq.6c) A - B = 2×E - D
Hint #7
In eq.6c, replace 2×E with A + F (from eq.1a): A - B = A + F - D In the equation above, subtract A from both sides, and add B and D to both sides: A - B - A + B + D = A + F - D - A + B + D which simplifies to eq.6d) D = F + B
Hint #8
In eq.2b, substitute F + B for D (from eq.6d): 5×E - 5×F = F + B - B which becomes 5×E - 5×F = F Add 5×F to both sides of the equation above: 5×E - 5×F + 5×F = F + 5×F which makes 5×E = 6×F Divide both sides by 5: 5×E ÷ 5 = 6×F ÷ 5 which makes E = 1⅕×F
Hint #9
Substitute (1⅕×F) for E in eq.1a: A + F = 2×(1⅕×F) which becomes A + F = 2⅖×F Subtract F from each side of the equation above: A + F - F = 2⅖×F - F which makes A = 1⅖×F
Hint #10
Substitute 1⅖×F for A in eq.6a: C × F = 1⅖×F × D Since F ≠ 0 (from eq.6), divide both sides of the above equation by F: (C × F) ÷ F = (1⅖×F × D) ÷ F which becomes eq.6e) C = 1⅖×D
Hint #11
Substitute (1⅕×F) for E, and 1⅖×F for A and C in eq.2a: 4×(1⅕×F) = 1⅖×F + 1⅖×D + D + F which becomes 4⅘×F = 2⅖×F + 2⅖×D Subtract 2⅖×F from each side of the above equation: 4⅘×F - 2⅖×F = 2⅖×F + 2⅖×D - 2⅖×F which makes 2⅖×F = 2⅖×D Divide both sides by 2⅖: 2⅖×F ÷ 2⅖ = 2⅖×D ÷ 2⅖ which makes F = D
Hint #12
Substitute F for D in eq.6e: C = 1⅖×F
Hint #13
Substitute 1⅖×F for A and C, 1⅕×F for E, and F for D in eq.4: 1⅖×F ÷ 1⅖×F = 1⅕×F - F which becomes 1 = ⅕×F Multiply both sides of the above equation by 5: 5 × 1 = 5 × ⅕×F which makes 5 = F and also makes 5 = F = D and makes A = C = 1⅖×F = 1⅖ × 5 = 7 E = 1⅕×F = 1⅕ × 5 = 6
Solution
Substitute 5 for D and F in eq.6d: 5 = 5 + B Subtract 5 from each side of the equation above: 5 - 5 = 5 + B - 5 which means 0 = B and makes ABCDEF = 707565