Puzzle for July 14, 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
Add C and E to both sides of eq.4: D – C + C + E = A + C – E + C + E which becomes eq.4a) D + E = A + 2×C
Hint #2
eq.6 may be written as: C = (A + D + E) ÷ 3 Multiply both sides of the above equation by 3: 3 × C = 3 × (A + D + E) ÷ 3 which becomes eq.6a) 3×C = A + D + E
Hint #3
In eq.6a, replace D + E with A + 2×C (from eq.4a): 3×C = A + A + 2×C which becomes 3×C = 2×A + 2×C Subtract 2×C from each side of the equation above: 3×C – 2×C = 2×A + 2×C – 2×C which makes C = 2×A
Hint #4
In eq.3, replace C with 2×A: 2×A – A = A – F which becomes A = A – F Subtract A from both sides of the equation above: A – A = A – F – A which makes 0 = –F which means 0 = F
Hint #5
In eq.2, substitute 0 for F: B + 0 = A + D which becomes eq.2a) B = A + D
Hint #6
Substitute 2×A for C, and 0 for F in eq.5: A + B = 2×A + E – 0 which becomes A + B = 2×A + E Subtract A from each side of the above equation: A + B – A = 2×A + E – A which becomes eq.5a) B = A + E
Hint #7
Substitute A + E for B (from eq.5a) into eq.2a: A + E = A + D Subtract A from each side of the equation above: A + E – A = A + D – A which makes E = D
Hint #8
Substitute D for E, and (2×A) for C in eq.4a: D + D = A + 2×(2×A) which becomes 2×D = A + 4×A which makes 2×D = 5×A Divide both sides of the above equation by 2: 2×D ÷ 2 = 5×A ÷ 2 which makes D = 2½×A and also makes E = D = 2½×A
Hint #9
Substitute 2½×A for D in eq.2a: B = A + 2½×A which makes B = 3½×A
Solution
Substitute 3½×A for B, 2×A for C, 2½×A for D and E, and 0 for F in eq.1: A + 3½×A + 2×A + 2½×A + 2½×A + 0 = 23 which simplifies to 11½×A = 23 Divide both sides of the above equation by 11½: 11½×A ÷ 11½ = 23 ÷ 11½ which means A = 2 making B = 3½×A = 3½ × 2 = 7 C = 2×A = 2 × 2 = 4 D = E = 2½×A = 2½ × 2 = 5 and ABCDEF = 274550