Puzzle for October 21, 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
eq.5 may be written as: F = (A + C + E) ÷ 3 Multiply both sides of the above equation by 3: 3 × F = 3 × (A + C + E) ÷ 3 which becomes eq.5a) 3×F = A + C + E
Hint #2
Add C to both sides of eq.4: C + F + C = A + E + C which is equivalent to eq.4a) 2×C + F = A + C + E
Hint #3
In eq.4a, replace A + C + E with 3×F (from eq.5a): 2×C + F = 3×F Subtract F from each side of the above equation: 2×C + F – F = 3×F – F which makes 2×C = 2×F Divide both sides by 2: 2×C ÷ 2 = 2×F ÷ 2 which makes C = F
Hint #4
In eq.6, substitute C for F: D = E + (C ÷ C) which makes eq.6a) D = E + 1
Hint #5
In eq.2, substitute C for F: B + C = C Subtract C from each side of the above equation: B + C – C = C – C which means B = 0
Hint #6
Substitute 0 for B in eq.3: E + F = A + 0 which makes eq.3a) E + F = A
Hint #7
Substitute F for C, and E + F for A (from eq.3a) in eq.4: F + F = E + F + E which becomes 2×F = 2×E + F Subtract F from each side of the equation above: 2×F – F = 2×E + F – F which makes F = 2×E and also makes C = F = 2×E
Hint #8
Substitute 2×E for F in eq.3a: E + 2×E = A which makes 3×E = A
Solution
Substitute 3×E for A, 0 for B, 2×E for C and F, and E + 1 for D (from eq.6a) in eq.1: 3×E + 0 + 2×E + E + 1 + E + 2×E = 28 which simplifies to 9×E + 1 = 28 Subtract 1 from both sides of the equation above: 9×E + 1 – 1 = 28 – 1 which makes 9×E = 27 Divide by both sides by 9: 9×E ÷ 9 = 27 ÷ 9 which means E = 3 making A = 3×E = 3 × 3 = 9 C = F = 2×E = 2 × 3 = 6 D = E + 1 = 3 + 1 = 4 (from eq.6a) and ABCDEF = 906436