Puzzle for March 7, 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
Subtract C from both sides of eq.5: D – A + E – C = A + C – E – C which becomes eq.5a) D – A + E – C = A – E In eq.2, replace A – E with D – A + E – C (from eq.5a): C – A = D – A + E – C + F Add A and C to both sides of the above equation: C – A + A + C = D – A + E – C + F + A + C which becomes eq.2a) 2×C = D + E + F
Hint #2
Add C and E to both sides of eq.3: D – C – B + C + E = C – E + C + E which becomes eq.3a) D – B + E = 2×C In eq.2a, replace 2×C with D – B + E (from eq.3a): D – B + E = D + E + F Subtract D and E from each side of the equation above: D – B + E – D – E = D + E + F – D – E which simplifies to –B = F Since B and F are non-negative integers, the above equation makes: B = F = 0
Hint #3
In eq.2, substitute 0 for F: C – A = A – E + 0 Add A and E to each side of the equation above: C – A + A + E = A – E + 0 + A + E which becomes eq.2b) C + E = 2×A
Hint #4
Substitute 0 for B in eq.4: 0 + D = A – 0 + C – D + E Add D to each side of the above equation: 0 + D + D = A – 0 + C – D + E + D which becomes eq.4a) 2×D = A + C + E
Hint #5
Substitute 2×A for C + E (from eq.2b) in eq.4a: 2×D = A + 2×A which makes 2×D = 3×A Divide both sides of the above equation by 2: 2×D ÷ 2 = 3×A ÷ 2 which makes D = 1½×A
Hint #6
Substitute 1½×A for D in eq.5: 1½×A – A + E = A + C – E In the above equation, add E to each side, and subtract A from each side: 1½×A – A + E + E – A = A + C – E + E – A which becomes eq.5b) 2×E – ½×A = C
Hint #7
Substitute 2×E – ½×A for C (from eq.5b) in eq.2b: 2×E – ½×A + E = 2×A which becomes 3×E – ½×A = 2×A Add ½×A to each side of the equation above: 3×E – ½×A + ½×A = 2×A + ½×A which makes 3×E = 2½×A Divide both sides by 3: 3×E ÷ 3 = 2½×A ÷ 3 which makes E = ⅚×A
Hint #8
Substitute (⅚×A) for E in eq.5b: 2×(⅚×A) – ½×A = C which becomes 1⅔×A – ½×A = C which makes 1⅙×A = C
Solution
Substitute 0 for B and F, 1⅙×A for C, 1½×A for D, and ⅚×A for E in eq.1: A + 0 + 1⅙×A + 1½×A + ⅚×A + 0 = 27 which simplifies to 4½×A = 27 Divide both sides of the equation above by 4½: 4½×A ÷ 4½ = 27 ÷ 4½ which means A = 6 making C = 1⅙×A = 1⅙ × 6 = 7 D = 1½×A = 1½ × 6 = 9 E = ⅚×A = ⅚ × 6 = 5 and ABCDEF = 607950