Puzzle for March 19, 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.
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Hint #1
In eq.5, replace B + C with A – B + E: A – B + E – E = E + F which becomes eq.5a) A – B = E + F
Hint #2
Add A and F to both sides of eq.2: E – A + A + F = B – F + A + F which becomes E + F = B + A In the above equation, replace E + F with A – B (from eq.5a): A – B = A + B Add B to both sides, and subtract A from both sides: A – B + B – A = A + B + B – A which becomes 0 = 2×B which means 0 = B
Hint #3
In eq.4, substitute 0 for B: A – 0 + E = 0 + C which becomes eq.4a) A + E = C
Hint #4
Substitute A + E for C (from eq.4a) into eq.6: A + E + D – A – E – F = A + F which becomes D – F = A + F Add F to each side of the above equation: D – F + F = A + F + F which becomes eq.6a) D = A + 2×F
Hint #5
Substitute (A + E) for C (from eq.4a), and A + 2×F for D (from eq.6a) in eq.3: (A + E) – F = A – (A + E) + A + 2×F which becomes A + E – F = 2×A – A – E + 2×F which becomes A + E – F = A – E + 2×F In the equation above, subtract A from both sides, and add F and E to both sides: A + E – F – A + F + E = A – E + 2×F – A + F + E which simplifies to 2×E = 3×F Divide both sides by 2: 2×E ÷ 2 = 3×F ÷ 2 which makes E = 1½×F
Hint #6
Substitute 0 for B, and 1½×F for E in eq.5a: A – 0 = 1½×F + F which makes A = 2½×F
Hint #7
Substitute 2½×F for A in eq.6a: D = 2½×F + 2×F which makes D = 4½×F
Hint #8
Substitute 2½×F for A, and 1½×F for E in eq.4a: 2½×F + 1½×F = C which makes 4×F = C
Solution
Substitute 2½×F for A, 0 for B, 4×F for C, 4½×F for D, and 1½×F for E in eq.1: 2½×F + 0 + 4×F + 4½×F + 1½×F + F = 27 which simplifies to 13½×F = 27 Divide both sides of the equation above by 13½: 13½×F ÷ 13½ = 27 ÷ 13½ which means F = 2 making A = 2½×F = 2½ × 2 = 5 C = 4×F = 4 × 2 = 8 D = 4½×F = 4½ × 2 = 9 E = 1½×F = 1½ × 2 = 3 and ABCDEF = 508932