Puzzle for September 16, 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.
* EF is a 2-digit number (not E×F).
Scratchpad
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Hint #1
eq.6 may be written as: B = (C + E + F) ÷ 3 Multiply both sides of the above equation by 3: 3 × B = 3 × (C + E + F) ÷ 3 which becomes eq.6a) 3×B = C + E + F
Hint #2
In eq.4, subtract E from both sides, and add F to both sides: C - E + F - E + F = B + E - F - E + F which becomes eq.4a) C - 2×E + 2×F = B
Hint #3
In eq.6a, substitute (C - 2×E + 2×F) for B (from eq.4a): 3×(C - 2×E + 2×F) = C + E + F which becomes 3×C - 6×E + 6×F = C + E + F In the equation above, add 6×E to both sides, and subtract C and F from both sides: 3×C - 6×E + 6×F + 6×E - C - F = C + E + F + 6×E - C - F which becomes eq.6b) 2×C + 5×F = 7×E
Hint #4
Subtract F from each side of eq.2: A + F - F = D - F - F which becomes eq.2a) A = D - 2×F Subtract F from each side of eq.3: D + F - F = C + E - F - F which becomes eq.3a) D = C + E - 2×F
Hint #5
In eq.2a, replace D with C + E - 2×F (from eq.3a): A = C + E - 2×F - 2×F which becomes eq.2b) A = C + E - 4×F
Hint #6
eq.5 may be written as: 10×E + F = A + D + E In the above equation, replace A with C + E - 4×F (from eq.2b), and D with C + E - 2×F (from eq.3a): 10×E + F = C + E - 4×F + C + E - 2×F + E which becomes 10×E + F = 2×C + 3×E - 6×F Subtract F and 3×E from both sides of the equation above: 10×E + F - F - 3×E = 2×C + 3×E - 6×F - F - 3×E which becomes eq.5a) 7×E = 2×C - 7×F
Hint #7
Substitute 2×C - 7×F for 7×E (from eq.5a) in eq.6b: 2×C + 5×F = 2×C - 7×F In the above equation, subtract 2×C from both sides, and add 7×F to both sides: 2×C + 5×F - 2×C + 7×F = 2×C - 5×F - 2×C + 7×F which makes 12×F = 0 which means F = 0
Hint #8
Substitute 0 for F in eq.2a: A = D - 2×0 which makes A = D - 0 which makes A = D
Hint #9
Substitute 0 for F in eq.5a: 7×E = 2×C - 7×0 which makes 7×E = 2×C - 0 which makes 7×E = 2×C Divide both sides of the above equation by 2: 7×E ÷ 2 = 2×C ÷ 2 which makes 3½×E = C
Hint #10
Substitute 3½×E for C, and 0 for F in eq.2b: A = 3½×E + E - 4×0 which becomes A = 4½×E - 0 which makes A = 4½×E and also makes A = D = 4½×E
Hint #11
Substitute 3½×E for C, and 0 for F in eq.4a: 3½×E - 2×E + 2×0 = B which becomes 1½×E + 0 = B which makes 1½×E = B
Solution
Substitute 4½×E for A and D, 1½×E for B, 3½×E for C, and 0 for F in eq.1: 4½×E + 1½×E + 3½×E + 4½×E + E + 0 = 30 which simplifies to 15×E = 30 Divide both sides of the above equation by 15: 15×E ÷ 15 = 30 ÷ 15 which means E = 2 making A = D = 4½×E = 4½ × 2 = 9 B = 1½×E = 1½ × 2 = 3 C = 3½×E = 3½ × 2 = 7 making ABCDEF = 937920