Puzzle for October 18, 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
eq.6 may be written as: A + E = (B + C + D + F) ÷ 4 Multiply both sides of the above equation by 4: 4 × (A + E) = 4 × (B + C + D + F) ÷ 4 which becomes eq.6a) 4×A + 4×E = B + C + D + F
Hint #2
eq.1 may be written as: A + E + B + C + D + F = 25 In the above equation, replace B + C + D + F with 4×A + 4×E (from eq.6a): A + E + 4×A + 4×E = 25 which becomes 5×A + 5×E = 25 Divide both sides by 5: (5×A + 5×E) ÷ 5 = 25 ÷ 5 which becomes eq.1a) A + E = 5
Hint #3
Subtract E from both sides of eq.1a: A + E – E = 5 – E which becomes A = 5 – E In eq.4, substitute 5 – E for A: F – B = 5 – E + B Add B and E to both sides of the equation above: F – B + B + E = 5 – E + B + B + E which becomes F + E = 5 + 2×B which may be written as eq.4a) E + F = 5 + 2×B
Hint #4
Substitute 5 + 2×B for E + F (from eq.4a) in eq.3: eq.3a) 5 + 2×B = C + D
Hint #5
Substitute 5 + 2×B for C + D (from eq.3a) and for E + F (from eq.4a) in eq.1: A + B + 5 + 2×B + 5 + 2×B = 25 which becomes A + 5×B + 10 = 25 Subtract both 5×B and 10 from each side of the equation above: A + 5×B + 10 – 5×B – 10 = 25 – 5×B – 10 which makes eq.1b) A = 15 – 5×B
Hint #6
Substitute 15 – 5×B for A (from eq.1b) in eq.1a: 15 – 5×B + E = 5 In the above equation, subtract 15 from both sides, and add 5×B to both sides: 15 – 5×B + E – 15 + 5×B = 5 – 15 + 5×B which becomes E = –10 + 5×B which may be written as eq.1c) E = 5×B – 10
Hint #7
Substitute 15 – 5×B for A (from eq.1b) in eq.4: F – B = 15 – 5×B + B which becomes F – B = 15 – 4×B Add B to both sides of the equation above: F – B + B = 15 – 4×B + B which becomes eq.4b) F = 15 – 3×B
Hint #8
Subtract E from both sides of eq.2: D + E – E = A + B + C – E which becomes D = A + B + C – E Substitute A + B + C – E for D in eq.3: E + F = C + A + B + C – E which becomes E + F = A + B + 2×C – E In the above equation, subtract A and B from each side, and add E to each side: E + F – A – B + E = A + B + 2×C – E – A – B + E which simplifies to eq.3b) 2×E + F – A – B = 2×C
Hint #9
Substitute (5×B – 10) for E (from eq.1c), 15 – 3×B for F (from eq.4b), and (15 – 5×B) for A (from eq.1b) in eq.3b: 2×(5×B – 10) + 15 – 3×B – (15 – 5×B) – B = 2×C which becomes 10×B – 20 + 15 – 3×B – 15 + 5×B – B = 2×C which becomes 11×B – 20 = 2×C Divide both sides of the equation above by 2: (11×B – 20) ÷ 2 = 2×C ÷ 2 which makes eq.3c) 5½×B – 10 = C
Hint #10
Substitute 5½×B – 10 for C (from eq.3c), 5×B – 10 for E (from eq.1c), 15 – 3×B for F (from eq.4b), and (15 – 5×B) for A (from eq.1b) in eq.5: 5½×B – 10 + 5×B – 10 + 15 – 3×B = (15 – 5×B) × B which becomes 7½×B – 5 = 15×B – 5×B² In the equation above, subtract 15×B from both sides, and add 5×B² to both sides: 7½×B – 5 – 15×B + 5×B² = 15×B – 5×B² – 15×B + 5×B² which becomes –7½×B – 5 + 5×B² = 0 which may be written as 5×B² – 7½×B – 5 = 0 Divide both sides by 2½: (5×B² – 7½×B – 5) ÷ 2½ = 0 ÷ 2½ which becomes eq.5a) 2×B² – 3×B – 2 = 0
Hint #11
eq.5a is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for B in eq.5a yields: B = { (–1)×(–3) ± sq.rt.[(–3)² – (4 × (2) × (–2))] } ÷ (2 × 2) which becomes B = {3 ± sq.rt.(9 – (–16))} ÷ 4 which becomes B = {3 ± sq.rt.(25)} ÷ 4 which becomes B = {3 ± 5} ÷ 4 In the above equation, either B = (3 + 5) ÷ 4 = 8 ÷ 4 = 2 or B = (3 – 5) ÷ 4 = –2 ÷ 4 = –½ Since B must be a non-negative integer, then B ≠ –½ and therefore makes B = 2
Hint #12
Substitute 2 for B in eq.1b: A = 15 – 5×2 which becomes A = 15 – 10 which makes A = 5
Hint #13
Substitute 2 for B in eq.3c: 5½×2 – 10 = C which becomes 11 – 10 = C which makes 1 = C
Hint #14
Substitute 2 for B in eq.4b: F = 15 – 3×2 which becomes F = 15 – 6 which makes F = 9
Hint #15
Substitute 2 for B in eq.1c: E = 5×2 – 10 which becomes E = 10 – 10 which makes E = 0
Solution
Substitute 0 for E, 5 for A, 2 for B, and 1 for C in eq.2: D + 0 = 5 + 2 + 1 which makes D = 8 and ABCDEF = 521809