Puzzle for December 15, 2019 ( )
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
Subtract E from both sides of eq.4: C – D + E – E = A + D – E which becomes eq.4a) C – D = A + D – E In eq.3, add C to both sides, and subtract A from each side: B – C + C – A = A – D + C – A which becomes B – A = –D + C which is the same as B – A = C – D In the above equation, replace C – D with A + D – E (from eq.4a): B – A = A + D – E Add A and E to each side: B – A + A + E = A + D – E + A + E which becomes eq.3a) B + E = 2×A + D
Hint #2
Subtract E from both sides of eq.2: D + E – E = A + B – E which becomes eq.2a) D = A + B – E In eq.3a, replace D with A + B – E (from eq.2a): B + E = 2×A + A + B – E which becomes B + E = 3×A + B – E In the equation above, add E to each side, and subtract B from each side: B + E + E – B = 3×A + B – E + E – B which simplifies to 2×E = 3×A Divide each side by 2: 2×E ÷ 2 = 3×A ÷ 2 which becomes E = 1½×A
Hint #3
In eq.2, substitute 1½×A for E: D + 1½×A = A + B Subtract A and D from both sides of the above equation: D + 1½×A – A – D = A + B – A – D which becomes ½×A = B – D Multiply both sides by 2: 2 × ½×A = 2 × (B – D) which makes eq.2a) A = 2×B – 2×D
Hint #4
In eq.3, add C and D to both sides, and subtract A from both sides: B – C + C + D – A = A – D + C + D – A which becomes eq.3b) B + D – A = C Substitute B + D – A for C (from eq.3b) in eq.5: B + B + D – A – A – F = A + F which becomes 2×B + D – 2×A – F = A + F Add 2×A and F to both sides of the equation above: 2×B + D – 2×A – F + 2×A + F = A + F + 2×A + F which becomes eq.5a) 2×B + D = 3×A + 2×F
Hint #5
Substitute (2×B – 2×D) for A (from eq.2a) in eq.5a: 2×B + D = 3×(2×B – 2×D) + 2×F which becomes 2×B + D = 6×B – 6×D + 2×F In the equation above, subtract 6×B from both sides, and add 6×D to each side: 2×B + D – 6×B + 6×D = 6×B – 6×D + 2×F – 6×B + 6×D which becomes 7×D – 4×B = 2×F Divide both sides by 2: (7×D – 4×B) ÷ 2 = 2×F ÷ 2 which becomes eq.5b) 3½×D – 2×B = F
Hint #6
Substitute 3½×D – 2×B for F (from eq.5b), and 1½×A for E in eq.6: D + 3½×D – 2×B = 1½×A which becomes eq.6a) 4½×D – 2×B = 1½×A
Hint #7
Substitute (2×B – 2×D) for A (from eq.2a) in eq.6a: 4½×D – 2×B = 1½×(2×B – 2×D) which becomes 4½×D – 2×B = 3×B – 3×D Add 2×B and 3×D to both sides: 4½×D – 2×B + 2×B + 3×D = 3×B – 3×D + 2×B + 3×D which means 7½×D = 5×B Divide both sides by 5: 7½×D ÷ 5 = 5×B ÷ 5 which makes 1½×D = B
Hint #8
Substitute (1½×D) for B in eq.2a: A = 2×(1½×D) – 2×D which becomes A = 3×D – 2×D which makes A = D and also makes E = 1½×A = 1½×D
Hint #9
Substitute (1½×D) for B in eq.5b: 3½×D – 2×(1½×D) = F which becomes 3½×D – 3×D = F which makes ½×D = F
Hint #10
Substitute 1½×D for B, and A for D in eq.3b: 1½×D + D – D = C which makes 1½×D = C
Solution
Substitute D for A, 1½×D for B and C and E, and ½×D for F in eq.1: D + 1½×D + 1½×D + D + 1½×D + ½×D = 42 which simplifies to 7×D = 42 Divide both sides of the equation above by 7: 7×D ÷ 7 = 42 ÷ 7 which means D = 6 making A = D = 6 B = C = E = 1½×D = 1½ × 6 = 9 F = ½×D = ½ × 6 = 3 and ABCDEF = 699693