Puzzle for May 10, 2019 ( )
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.6 may be written as: A + E + C = D + F In the equation above, replace A + E with B + C + F (from eq.2): B + C + F + C = D + F Subtract F from both sides: B + C + F + C - F = D + F - F which becomes eq.6a) B + 2×C = D
Hint #2
Add both A and C to each side of eq.4: D - A + A + C = A + B - C + A + C which becomes D + C = 2×A + B In the above equation, replace D with B + 2×C (from eq.6a): B + 2×C + C = 2×A + B Subtract B from both sides: B + 2×C + C - B = 2×A + B - B which simplifies to 3×C = 2×A Divide both sides by 3: 3×C ÷ 3 = 2×A ÷ 3 which makes C = ⅔×A
Hint #3
Subtract both A and F from each side of eq.2: B + C + F - A - F = A + E - A - F which becomes B + C - A = E - F In the above equation, substitute A - D for E - F (from eq.3): B + C - A = A - D Add A + D to both sides of the above equation: B + C - A + A + D = A - D + A + D which becomes eq.2a) B + C + D = 2×A
Hint #4
Substitute B + 2×C for D (from eq.6a) in eq.2a: B + C + B + 2×C = 2×A which becomes 2×B + 3×C = 2×A Substitute (⅔×A) for C in the equation above: 2×B + 3×(⅔×A) = 2×A which becomes 2×B + 2×A = 2×A Subtract 2×A from both sides: 2×B + 2×A - 2×A = 2×A - 2×A which becomes 2×B = 0 which means B = 0
Hint #5
Substitute 0 for B, and (⅔×A) for C in eq.6a: 0 + 2×(⅔×A) = D which makes 1⅓×A = D
Hint #6
Substitute 1⅓×A for D in eq.3: E - F = A - 1⅓×A which becomes E - F = -⅓×A Add ⅓×A and F to both sides: E - F + ⅓×A + F = -⅓×A + ⅓×A + F which becomes eq.3a) E + ⅓×A = F
Hint #7
Substitute E + ⅓×A for F (from eq.3a), and 1⅓×A for D in eq.5: E + ⅓×A - 1⅓×A = 1⅓×A - E which becomes E - A = 1⅓×A - E Add A and E to both sides: E - A + A + E = 1⅓×A - E + A + E which simplifies to 2×E = 2⅓×A Divide both sides by 2: 2×E ÷ 2 = 2⅓×A ÷ 2 which makes E = 1⅙×A
Hint #8
Substitute 1⅙×A for E in eq.3a: 1⅙×A + ⅓×A = F which makes 1½×A = F
Solution
Substitute 0 for B, ⅔×A for C, 1⅓×A for D, 1⅙×A for E, and 1½×A for F in eq.1: A + 0 + ⅔×A + 1⅓×A + 1⅙×A + 1½×A = 34 which simplifies to 5⅔×A = 34 Divide both sides by 5⅔: 5⅔×A ÷ 5⅔ = 34 ÷ 5⅔ which means A = 6 making C = ⅔×A = ⅔ × 6 = 4 D = 1⅓×A = 1⅓ × 6 = 8 E = 1⅙×A = 1⅙ × 6 = 7 F = 1½×A = 1½ × 6 = 9 and ABCDEF = 604879