Puzzle for January 27, 2021 ( )
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 + D + E) ÷ 4 Multiply both sides of the above equation by 4: 4 × F = 4 × (A + C + D + E) ÷ 4 which becomes eq.5a) 4×F = A + C + D + E
Hint #2
In eq.5a, replace A + C with D (from eq.2): 4×F = D + D + E which becomes eq.5b) 4×F = 2×D + E Add E to both sides of eq.4: E + F + E = D – E + E which becomes eq.4a) 2×E + F = D
Hint #3
In eq.5b, substitute (2×E + F) for D (from eq.4a): 4×F = 2×(2×E + F) + E which is equivalent to 4×F = 4×E + 2×F + E which becomes 4×F = 5×E + 2×F Subtract 2×F from each side of the equation above: 4×F – 2×F = 5×E + 2×F – 2×F which makes 2×F = 5×E Divide both sides by 2: 2×F ÷ 2 = 5×E ÷ 2 which makes F = 2½×E
Hint #4
Substitute 2½×E for F in eq.4a: 2×E + 2½×E = D which makes 4½×E = D
Hint #5
Substitute 4½×E for D, and 2½×E for F in eq.6: A = (4½×E + 2½×E) ÷ E which becomes A = (7×E) ÷ E which makes A = 7
Hint #6
Substitute 4½×E for D, and 7 for A in eq.2: 4½×E = 7 + C Subtract 7 from each side of the above equation: 4½×E – 7 = 7 + C – 7 which makes eq.2a) 4½×E – 7 = C
Hint #7
Substitute 4½×E for D, 4½×E – 7 for C (from eq.2a), and 2½×E for F in eq.3: B + 4½×E = 4½×E – 7 + E + 2½×E – B which becomes B + 4½×E = 8×E – 7 – B In the equation above, subtract 4½×E from both sides, and add B to both sides: B + 4½×E – 4½×E + B = 8×E – 7 – B – 4½×E + B which becomes 2×B = 3½×E – 7 Divide both sides by 2: 2×B ÷ 2 = (3½×E – 7) ÷ 2 which makes eq.3a) B = 1¾×E – 3½
Solution
Substitute 7 for A, 1¾×E – 3½ for B (from eq.3a), 4½×E – 7 for C (from eq.2a), 4½×E for D, and 2½×E for F in eq.1: 7 + 1¾×E – 3½ + 4½×E – 7 + 4½×E + E + 2½×E = 25 which simplifies to 14¼×E – 3½ = 25 Add 3½ to both sides of the equation above: 14¼×E – 3½ + 3½ = 25 + 3½ which makes 14¼×E = 28½ Divide both sides by 14¼: 14¼×E ÷ 14¼ = 28½ ÷ 14¼ which means E = 2 making B = 1¾×E – 3½ = 1¾×2 – 3½ = 3½ – 3½ = 0 (from eq.3a) C = 4½×E – 7 = 4½×2 – 7 = 9 – 7 = 2 (from eq.2a) D = 4½×E = 4½ × 2 = 9 F = 2½×E = 2½ × 2 = 5 and ABCDEF = 702925