Puzzle for February 20, 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
In eq.4, replace D with E + F (from eq.1): E + F - F = A - E which becomes E = A - E Add E to each side of the above equation: E + E = A - E + E which makes eq.4a) 2×E = A
Hint #2
In eq.3, replace A with 2×E: C - E = 2×E - D Add D and E to both sides of the above equation: C - E + D + E = 2×E - D + D + E which becomes eq.3a) C + D = 3×E
Hint #3
Add C to both sides of eq.1: E + F + C = D + C which may be written as E + F + C = C + D In the above equation, substitute 3×E for C + D (from eq.3a): E + F + C = 3×E Subtract E from each side: E + F + C - E = 3×E - E which becomes F + C = 2×E which may be written as eq.1a) C + F = 2×E
Hint #4
In eq.2, substitute 2×E for A, and 2×E for C + F (from eq.1a): 2×E + B = 2×E Subtract 2×E from each side of the above equation: 2×E + B - 2×E = 2×E - 2×E which makes B = 0
Hint #5
In eq.5, replace B with 0: F - 0 = C - F Add F to both sides of the above equation: F - 0 + F = C - F + F which makes 2×F = C
Hint #6
Substitute 2×F for C in eq.1a: 2×F + F = 2×E which means 3×F = 2×E Divide both sides by 2: 3×F ÷ 2 = 2×E ÷ 2 which makes 1½×F = E
Hint #7
Substitute (1½×F) for E in eq.4a: 2×(1½×F) = A which makes 3×F = A
Hint #8
Substitute 1½×F for E in eq.1: 1½×F + F = D which makes 2½×F = D
Solution
Substitute 3×F for A, 2×F for C, and 2½×F for D in eq.6: (3×F + 2×F) ÷ F = 2½×F which is the same as 5×F ÷ F = 2½×F which means 5 = 2½×F Divide both sides by 2½: 5 ÷ 2½ = 2½×F ÷ 2½ which means 2 = F making A = 3×F = 3 × 2 = 6 C = 2×F = 2 × 2 = 4 D = 2½×F = 2½ × 2 = 5 E = 1½×F = 1½ × 2 = 3 and ABCDEF = 604532