Puzzle for July 31, 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.4 may be written as: F = (A + C + D) ÷ 3 Multiply both sides of the above equation by 3: 3 × F = 3 × (A + C + D) ÷ 3 which becomes eq.4a) 3×F = A + C + D
Hint #2
In eq.2, replace A + C + D with 3×F (from eq.4a): eq.2a) B + E = 3×F – E Subtract E from each side of eq.2a: B + E – E = 3×F – E – E which makes eq.2b) B = 3×F – 2×E
Hint #3
eq.1 may be written as: A + C + D + B + E + F = 33 In the above equation, replace A + C + D with 3×F (from eq.4a), and B + E with 3×F – E (from eq.2a): 3×F + 3×F – E + F = 33 which becomes 7×F – E = 33 Add E to both sides, and subtract 33 from both sides: 7×F – E + E – 33 = 33 + E – 33 which becomes eq.1a) 7×F – 33 = E
Hint #4
To make eq.1a true, check several possible values for F and E: If F = 5, then E = 7×5 – 33 = 35 – 33 = 2 If F = 6, then E = 7×6 – 33 = 42 – 33 = 9 If F = 7, then E = 7×7 – 33 = 49 – 33 = 16 If F > 7, then E > 16 If F = 4, then E = 7×4 – 33 = 28 – 33 = –5 If F < 4, then E < –5 Since E must be a one-digit non-negative integer, the above equations make: F = 5 and E = 2 or F = 6 and E = 9
Hint #5
Check: F = 5, and E = 2 ... Substituting 5 for F, and 2 for E in eq.2b would yield: B = 3×5 – 2×2 which would become B = 15 – 4 which would make B = 11 Since B must be a one-digit non-negative integer, then: B ≠ 11 which means F ≠ 5 and E ≠ 2 and therefore means F = 6 and E = 9
Hint #6
Substitute 6 for F, and 9 for E in eq.2b: B = 3×6 – 2×9 which becomes B = 18 – 18 which makes B = 0
Hint #7
Substitute 0 for B in eq.3: A + C – F = 0 – C + F which becomes A + C – F = –C + F Add F and C to both sides of the above equation: A + C – F + F + C = –C + F + F + C which becomes eq.3a) A + 2×C = 2×F
Hint #8
Substitute 0 for B in eq.5: F – (0 ÷ A) = (A ÷ C) + C which becomes F – 0 = (A ÷ C) + C which becomes F = (A ÷ C) + C Multiply both sides of the above equation by 2: 2×F = 2×((A ÷ C) + C) which becomes eq.5a) 2×F = 2×(A ÷ C) + 2×C
Hint #9
Substitute A + 2×C for 2×F (from eq.3a) into eq.5a: A + 2×C = 2×(A ÷ C) + 2×C Subtract 2×C from each side of the above equation: A + 2×C – 2×C = (2×A ÷ C) + 2×C – 2×C which becomes 2×A ÷ C = A Multiply both sides by C: C × 2×A ÷ C = C × A which becomes 2×A = C × A Since A ≠ 0 (from eq.5), divide both sides by A: 2×A ÷ A = (C × A) ÷ A which becomes 2 = C
Hint #10
Substitute 2 for C, and 6 for F in eq.3a: A + 2×2 = 2×6 which becomes A + 4 = 12 Subtract 4 from each side of the equation above: A + 4 – 4 = 12 – 4 which makes A = 8
Solution
Substitute 6 for F, 8 for A, and 2 for C in eq.4a: 3×6 = 8 + 2 + D which becomes 18 = 10 + D Subtract 10 from each side of the above equation: 18 – 10 = 10 + D – 10 which makes 8 = D and makes ABCDEF = 802896