Puzzle for April 3, 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.
Scratchpad
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Hint #1
In eq.6, replace B + E with C + F (from eq.2): C + D – B = C + F + F which becomes C + D – B = C + 2×F In the equation above, subtract C from both sides, and add B to both sides: C + D – B – C + B = C + 2×F – C + B which becomes eq.6a) D = 2×F + B
Hint #2
In eq.5, replace D with 2×F + B (from eq.6a): A + E + F = B + 2×F + B – A which becomes A + E + F = 2×B + 2×F – A In the equation above, subtract F from both sides, and add A to both sides: A + E + F – F + A = 2×B + 2×F – A – F + A which becomes eq.5a) 2×A + E = 2×B + F
Hint #3
In eq.4, replace D with 2×F + B (from eq.6a): B + C = A + 2×F + B + F which becomes B + C = A + 3×F + B Subtract B from each side of the equation above: B + C – B = A + 3×F + B – B which becomes eq.4a) C = A + 3×F
Hint #4
In eq.3, substitute A + 3×F for C (from eq.4a): A + 3×F + E = A + B Subtract A from each side of the above equation: A + 3×F + E – A = A + B – A which becomes eq.3a) 3×F + E = B
Hint #5
Substitute 3×F + E for B (from eq.3a) in eq.5a: 2×A + E = 2×(3×F + E) + F which becomes 2×A + E = 6×F + 2×E + F which becomes 2×A + E = 7×F + 2×E Subtract E from both sides of the above equation: 2×A + E – E = 7×F + 2×E – E which becomes 2×A = 7×F + E Divide both sides by 2: 2×A ÷ 2 = (7×F + E) ÷ 2 which becomes eq.5b) A = 3½×F + ½×E
Hint #6
Substitute 3½×F + ½×E for A (from eq.5b) in eq.4a: C = 3½×F + ½×E + 3×F which becomes eq.4b) C = 6½×F + ½×E
Hint #7
Substitute 3×F + E for B (from eq.3a), and 6½×F + ½×E for C (from eq.4b) into eq.2: 3×F + E + E = 6½×F + ½×E + F which becomes 3×F + 2×E = 7½×F + ½×E Subtract 3×F and ½×E from each side of the equation above: 3×F + 2×E – 3×F – ½×E = 7½×F + ½×E – 3×F – ½×E which makes 1½×E = 4½×F Divide both sides by 1½: 1½×E ÷ 1½ = 4½×F ÷ 1½ which makes E = 3×F
Hint #8
Substitute (3×F) for E in eq.5b: A = 3½×F + ½×(3×F) which becomes A = 3½×F + 1½×F which becomes A = 5×F
Hint #9
Substitute 3×F for E in eq.3a: 3×F + 3×F = B which makes 6×F = B
Hint #10
Substitute (3×F) for E in eq.4b: C = 6½×F + ½×(3×F) which becomes C = 6½×F + 1½×F which makes C = 8×F
Hint #11
Substitute 6×F for B in eq.6a: D = 2×F + 6×F which makes D = 8×F
Solution
Substitute 5×F for A, 6×F for B, 8×F for C and D, and 3×F for E in eq.1: 5×F + 6×F + 8×F + 8×F + 3×F + F = 31 which simplifies to 31×F = 31 Divide both sides of the above equation by 31: 31×F ÷ 31 = 31 ÷ 31 which means F = 1 making A = 5×F = 5 × 1 = 5 B = 6×F = 6 × 1 = 6 C = D = 8×F = 8 × 1 = 8 E = 3×F = 3 × 1 = 3 and ABCDEF = 568831