Puzzle for January 2, 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 positive integer.
Scratchpad
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Hint #1
In eq.5, replace C with B + D (from eq.2): D – E = B + D – A In the equation above, subtract D from both sides, and add A to both sides: D – E – D + A = B + D – A – D + A which becomes –E + A = B which may be written as eq.5a) A – E = B
Hint #2
In eq.4, replace B with A – E (from eq.5a): C + E = A + A – E – E which becomes C + E = 2×A – 2×E Subtract E from both sides of the equation above: C + E – E = 2×A – 2×E – E which becomes eq.4a) C = 2×A – 3×E
Hint #3
In eq.2, substitute A – E for B (from eq.5a), and 2×A – 3×E for C (from eq.4a): A – E + D = 2×A – 3×E In the equation above, subtract A from both sides, and add E to both sides: A – E + D – A + E = 2×A – 3×E – A + E which becomes eq.2a) D = A – 2×E
Hint #4
Substitute A – 2×E for D (from eq.2a) in eq.3: A – 2×E + E = F which becomes A – E = F In the equation above substitute B for A – E (from eq.5a): B = F which means eq.3a) B = F = A – E
Hint #5
Substitute (A – E) for B and F (from eq.3a), and (A – 2×E) for D (from eq.2a) in eq.6: (A – E) × (A – E) = (A × (A – 2×E)) + E which becomes A×A – A×E – E×A + E×E = A×A – A×2×E + E which is equivalent to A² – 2×A×E + E² = A² – 2×A×E + E In the equation above, subtract A² from each side, and add 2×A×E to each side: A² – 2×A×E + E² – A² + 2×A×E = A² – 2×A×E + E – A² + 2×A×E which simplifies to E² = E Divide both sides by E: E² ÷ E = E ÷ E which makes E = 1
Hint #6
Substitute A – E for B and F (from eq.3a), 2×A – 3×E for C (from eq.4a), and A – 2×E for D (from eq.2a) in eq.1: A + A – E + 2×A – 3×E + A – 2×E + E + A – E = 30 which simplifies to eq.1a) 6×A – 6×E = 30
Solution
Substitute 1 for E in eq.1a: 6×A – 6×1 = 30 which becomes 6×A – 6 = 30 Add 6 to both sides: 6×A – 6 + 6 = 30 + 6 which makes 6×A = 36 Divide both sides by 6: 6×A ÷ 6 = 36 ÷ 6 which means A = 6 making B = F = A – E = 6 – 1 = 5 (from eq.3a) C = 2×A – 3×E = 2×6 – 3×1 = 12 – 3 = 9 (from eq.4a) D = A – 2×E = 6 – 2×1 = 6 – 2 = 4 (from eq.2a) and ABCDEF = 659415