Puzzle for August 2, 2020 ( )
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.2, replace D + E with A + C – D (from eq.5): B + A + C – D = A + F Subtract A and C from both sides of the equation above: B + A + C – D – A – C = A + F – A – C which becomes eq.2a) B – D = F – C
Hint #2
Subtract A and D from both sides of eq.3: A + B – A – D = C + D – A – D which becomes B – D = C – A In eq.2a, replace B – D with C – A: C – A = F – C Add A and C to both sides of the equation above: C – A + A + C = F – C + A + C which becomes 2×C = F + A which may be written eq.2b) 2×C = A + F
Hint #3
In eq.2, substitute 2×C for A + F (from eq.2b): eq.2c) B + D + E = 2×C
Hint #4
eq.1 may be re-written as: A + F + B + D + E + C = 30 In the equation above, substitute 2×C for A + F (from eq.2b), and 2×C for B + D + E (from eq.2c): 2×C + 2×C + C = 30 which makes 5×C = 30 Divide both sides by 5: 5×C ÷ 5 = 30 ÷ 5 which makes C = 6
Hint #5
Substitute 6 for C in eq.2b: 2×6 = A + F which becomes 12 = A + F Subtract A from each side of the equation above: 12 – A = A + F – A which makes eq.2d) 12 – A = F
Hint #6
Subtract D from each side of eq.3: A + B – D = C + D – D which becomes eq.3a) A + B – D = C Subtract E from each side of eq.4: C + E – E = B + F – E which becomes eq.4a) C = B + F – E
Hint #7
In eq.4a, substitute A + B – D for C (from eq.3a): A + B – D = B + F – E Subtract B from each side of the above equation: A + B – D – B = B + F – E – B which becomes eq.4b) A – D = F – E
Hint #8
Subtract A and E from both sides of eq.2: B + D + E – A – E = A + F – A – E which becomes B + D – A = F – E Substitute A – D for F – E (from eq.4b) in the equation above: B + D – A = A – D Add A to each side, and subtract D from each side: B + D – A + A – D = A – D + A – D which becomes eq.2e) B = 2×A – 2×D
Hint #9
Substitute 2×A – 2×D for B (from eq.2e), 12 – A for F (from eq.2d), and 6 for C in eq.2a: 2×A – 2×D – D = 12 – A – 6 which becomes 2×A – 3×D = 6 – A Add A to both sides of the equation above: 2×A – 3×D + A = 6 – A + A which becomes 3×A – 3×D = 6 Divide both sides by 3: (3×A – 3×D) ÷ 3 = 6 ÷ 3 which becomes A – D = 2 Add D to both sides, and subtract 2 from both sides: A – D + D – 2 = 2 + D + 2 which makes eq.2f) A – 2 = D
Hint #10
Substitute (A – 2) for D in eq.2e: B = 2×A – 2×(A – 2) which becomes B = 2×A – 2×A + 4 which makes B = 4
Hint #11
Substitute 4 for B, and 6 for C in eq.6: 4 × F = 6 × E Divide both sides of the above equation by 4: (4 × F) ÷ 4 = (6 × E) ÷ 4 which makes F = 1½×E
Hint #12
Substitute 6 for C, 4 for B, and 1½×E for F in eq.4: 6 + E = 4 + 1½×E Subtract E and 4 from both sides of the equation above: 6 + E – E – 4 = 4 + 1½×E – E – 4 which means 2 = ½×E Divide both sides by ½: 2 ÷ ½ = ½×E ÷ ½ which makes 4 = E making F = 1½×E = 1½ × 4 = 6
Hint #13
Substitute 6 for F in eq.2d: 12 – A = 6 In the equation above, add A to both sides, and subtract 6 from both sides: 12 – A + A – 6 = 6 + A – 6 which means 6 = A
Solution
Substitute 4 for B, and 6 for F and C in eq.2a: 4 – D = 6 – 6 which becomes 4 – D = 0 Add D to both sides of the above equation: 4 – D + D = 0 + D which makes 4 = D and ABCDEF = 646446