Puzzle for January 7, 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
eq.6 may be written as: A + B – D = D + F + E – C In the above equation, replace D + F with A (from eq.2): A + B – D = A + E – C Subtract A from both sides, and add D and C to both sides: A + B – D – A + D + C = A + E – C – A + D + C which simplifies to eq.6a) B + C = E + D
Hint #2
Add C and E to both sides of eq.4: D – C + C + E = A – E + C + E which becomes D + E = A + C which is the same as eq.4a) E + D = A + C
Hint #3
In eq.4a, replace E + D with B + C (from eq.6a): B + C = A + C Subtract C from each side of the equation above: B + C – C = A + C – C which makes B = A
Hint #4
In eq.6, substitute A for B, and C + D for E + F (from eq.3): A + A – D = D + C + D – C which becomes 2×A – D = 2×D Add D to both sides of the above equation: 2×A – D + D = 2×D + D which becomes 2×A = 3×D Divide both sides by 2: 2×A ÷ 2 = 3×D ÷ 2 which makes A = 1½×D and also makes eq.6b) B = A = 1½×D
Hint #5
Substitute 1½×D for A (from eq.6b) in eq.2: 1½×D = D + F Subtract D from each side of the above equation: 1½×D – D = D + F – D which makes eq.2a) ½×D = F
Hint #6
Substitute ½×D for F (from eq.2a) in eq.3: C + D = E + ½×D Subtract ½×D from both sides of the equation above: C + D – ½×D = E + ½×D – ½×D which becomes eq.3a) C + ½×D = E
Hint #7
Substitute (C + ½×D) for E (from eq.3a), and 1½×D for A and B (from eq.6b) in eq.5: C + (C + ½×D) = 1½×D + 1½×D – (C + ½×D) which becomes 2×C + ½×D = 3×D – C – ½×D which becomes 2×C + ½×D = 2½×D – C In the above equation, subtract ½×D from both sides, and add C to both sides: 2×C + ½×D – ½×D + C = 2½×D – C – ½×D + C which makes 3×C = 2×D Divide both sides by 2: 3×C ÷ 2 = 2×D ÷ 2 which makes 1½×C = D
Hint #8
Substitute (1½×C) for D in eq.2a: ½×(1½×C) = F which makes ¾×C = F
Hint #9
Substitute (1½×C) for D in eq.6b: B = A = 1½×(1½×C) which makes B = A = 2¼×C
Hint #10
Substitute (1½×C) for D in eq.3a: C + ½×(1½×C) = E which becomes C + ¾×C = E which makes 1¾×C = E
Solution
Substitute 2¼×C for A and B, 1½×C for D, 1¾×C for E, and ¾×C for F in eq.1: 2¼×C + 2¼×C + C + 1½×C + 1¾×C + ¾×C = 38 which simplifies to 9½×C = 38 Divide both sides of the above equation by 9½: 9½×C ÷ 9½ = 38 ÷ 9½ which means C = 4 making A = B = 2¼×C = 2¼ × 4 = 9 D = 1½×C = 1½ × 4 = 6 E = 1¾×C = 1¾ × 4 = 7 F = ¾×C = ¾ × 4 = 3 and ABCDEF = 994673