Puzzle for December 27, 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
In eq.4, replace D with B + E (from eq.2): A + B = B + E + E which becomes A + B = B + 2×E Subtract B from each side of the equation above: A + B - B = B + 2×E - B which makes A = 2×E
Hint #2
In eq.5, replace A with 2×E, and C with E + F (from eq.3): D + F = 2×E + E + F which becomes D + F = 3×E + F Subtract F from each side of the above equation: D + F - F = 3×E + F - F which makes D = 3×E
Hint #3
In eq.2, substitute 3×E for D: 3×E = B + E Subtract E from both sides of the above equation: B + E - E = 3×E - E which makes B = 2×E
Hint #4
eq.6 may be written as: C = (A + B + E + F) ÷ 4 Multiply both sides of the above equation by 4: 4 × C = 4 × (A + B + E + F) ÷ 4 which becomes eq.6a) 4×C = A + B + E + F Subtract E from each side of eq.3: C - E = E + F - E which becomes eq.3a) C - E = F
Hint #5
In eq.6a, replace A and B with 2×E, and F with C - E (from eq.3a): 4×C = 2×E + 2×E + E + C - E which becomes 4×C = 4×E + C Subtract C from both sides of the equation above: 4×C - C = 4×E + C - C which makes 3×C = 4×E Divide both sides by 3: 3×C ÷ 3 = 4×E ÷ 3 which makes C = 1⅓×E
Hint #6
Substitute 1⅓×E for C in eq.3a: 1⅓×E - E = F which makes ⅓×E = F
Solution
Substitute 2×E for A and B, 1⅓×E for C, 3×E for D, and ⅓×E for F in eq.1: 2×E + 2×E + 1⅓×E + 3×E + E + ⅓×E = 29 which simplifies to 9⅔×E = 29 Divide both sides of the above equation by 9⅔: 9⅔×E ÷ 9⅔ = 29 ÷ 9⅔ which means E = 3 making A = B = 2×E = 2 × 3 = 6 C = 1⅓×E = 1⅓ × 3 = 4 D = 3×E = 3 × 3 = 9 F = ⅓×E = ⅓ × 3 = 1 and ABCDEF = 664931