Puzzle for November 25, 2023 ( )
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
Multiply both sides of eq.3 by (-1): (-1) × (C - E) = (-1) × (E - F) which becomes -C + E = -E + F which may be written as eq.3a) E - C = F - E
Hint #2
In eq.4, replace F - E with E - C (from eq.3a): E - C = C - D - F Add C, D, and F to both sides of the above equation: E - C + C + D + F = C - F - D + C + D + F which becomes E + D + F = 2×C which may be written eq.4a) D + E + F = 2×C
Hint #3
In eq.5, replace D + E + F with 2×C (from eq.4a): A + C = 2×C Subtract C from each side of the equation above: A + C - C = 2×C - C which makes A = C
Hint #4
eq.6 may be written as: E + F = (A + B + C + D) ÷ 4 Multiply both sides of the above equation by 4: 4 × (E + F) = 4 × (A + B + C + D) ÷ 4 which becomes eq.6a) 4×E + 4×F = A + B + C + D
Hint #5
In eq.1, substitute 4×E + 4×F for A + B + C + D (from eq.6a): 4×E + 4×F + E + F = 35 which becomes 5×E + 5×F = 35 Divide both sides of the above equation by 5: (5×E + 5×F) ÷ 5 = 35 ÷ 5 which becomes E + F = 7 Subtract E from each side: E + F - E = 7 - E which makes eq.1a) F = 7 - E
Hint #6
Substitute (7 - E) for F (from eq.1a) in eq.3: C - E = E - (7 - E) which becomes C - E = E - 7 + E which becomes C - E = 2×E - 7 Add E to both sides of the equation above: C - E + E = 2×E - 7 + E which makes C = 3×E - 7 and also makes eq.3b) A = C = 3×E - 7
Hint #7
eq.5 may be written as: A + C = E + D + F Substitute 3×E - 7 for C (from eq.3b), and A + B for D + F (from eq.2) in the above equation: A + 3×E - 7 = E + A + B Subtract A and E from both sides: A + 3×E - 7 - A - E = E + A + B - A - E which makes eq.5a) 2×E - 7 = B
Hint #8
Substitute 7 - E for F (from eq.1a), and (3×E - 7) for C (from eq.3b) in eq.4a: D + E + 7 - E = 2×(3×E - 7) which becomes D + 7 = 6×E - 14 Subtract 7 from each side of the above equation: D + 7 - 7 = 6×E - 14 - 7 which makes eq.4b) D = 6×E - 21
Hint #9
Substitute 3×E - 7 for A and C (from eq.3b), 2×E - 7 for B (from eq.5a), 6×E - 21 for D (from eq.4b), and 7 - E for F (from eq.1a) in eq.1: 3×E - 7 + 2×E - 7 + 3×E - 7 + 6×E - 21 + E + 7 - E = 35 which simplifies to 14×E - 35 = 35 Add 35 to both sides of the equation above: 14×E - 35 + 35 = 35 + 35 which makes 14×E = 70 Divide both sides by 14: 14×E ÷ 14 = 70 ÷ 14 which means E = 5
Solution
Since E = 5, then: A = C = 3×E - 7 = 3×5 - 7 = 15 - 7 = 8 (from eq.3b) B = 2×E - 7 = 2×5 - 7 = 10 - 7 = 3 (from eq.5a) D = 6×E - 21 = 6×5 - 21 = 30 - 21 = 9 (from eq.4b) F = 7 - E = 7 - 5 = 2 (from eq.1a) and ABCDEF = 838952