Puzzle for October 9, 2021 ( )
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.4, subtract B from both sides, and add A to both sides: A + B – B + A = C + D – A – B + A which becomes 2×A = C + D – B which may be written as 2×A = C – B + D In the above equation, replace C – B with D + E (from eq.3): 2×A = D + E + D which becomes eq.4a) 2×A = 2×D + E
Hint #2
Add B to both sides of eq.3: C – B + B = D + E + B which becomes C = D + E + B In eq.5, replace C with D + E + B: B + E + F = D + E + B + D – E which becomes B + E + F = 2×D + B Subtract B from each side of the equation above: B + E + F – B = 2×D + B – B which becomes eq.3a) E + F = 2×D
Hint #3
In eq.3a, substitute A + E for F (from eq.2): E + A + E = 2×D which becomes eq.3b) A + 2×E = 2×D
Hint #4
Substitute A + 2×E for 2×D (from eq.3b) into eq.4a: 2×A = A + 2×E + E which becomes 2×A = A + 3×E Subtract A from each side of the above equation: 2×A – A = A + 3×E – A which makes A = 3×E
Hint #5
Substitute 3×E for A in eq.2: F = 3×E + E which makes F = 4×E
Hint #6
Substitute 3×E for A in eq.3b: 3×E + 2×E = 2×D which makes 5×E = 2×D Divide both sides of the above equation by 2: 5×E ÷ 2 = 2×D ÷ 2 which makes 2½×E = D
Hint #7
eq.6 may be written as: D = (B + C + F) ÷ 3 Multiply both sides of the above equation by 3: 3 × D = 3 × (B + C + F) ÷ 3 which becomes eq.6a) 3×D = B + C + F
Hint #8
Substitute (2½×E) for D, and 4×E for F in eq.6a: 3×(2½×E) = B + C + 4×E which becomes 7½×E = B + C + 4×E Subtract B and 4×E from each side of the equation above: 7½×E – B – 4×E = B + C + 4×E – B – 4×E which becomes eq.6b) 3½×E – B = C
Hint #9
Substitute 3×E for A, 3½×E – B for C (from eq.6b), and 2½×E for D in eq.4: 3×E + B = 3½×E – B + 2½×E – 3×E which becomes 3×E + B = 3×E – B In the above equation, subtract 3×E from both sides, and add B to both sides: 3×E + B – 3×E + B = 3×E – B – 3×E + B which becomes 2×B = 0 which means B = 0
Hint #10
Substitute 0 for B in eq.6b: 3½×E – 0 = C which makes 3½×E = C
Solution
Substitute 3×E for A, 0 for B, 3½×E for C, 2½×E for D, and 4×E for F in eq.1: 3×E + 0 + 3½×E + 2½×E + E + 4×E = 28 which simplifies to 14×E = 28 Divide both sides of the above equation by 14: 14×E ÷ 14 = 28 ÷ 14 which means E = 2 making A = 3×E = 3 × 2 = 6 C = 3½×E = 3½ × 2 = 7 D = 2½×E = 2½ × 2 = 5 F = 4×E = 4 × 2 = 8 and ABCDEF = 607528