Puzzle for April 25, 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.
* DE is a 2-digit number (not D×E).
Scratchpad
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Hint #1
In eq.2, replace A + C with B + D (from eq.1): B + E = B + D + D which becomes B + E = B + 2×D Subtract B from both sides of the equation above: B + E – B = B + 2×D – B which makes E = 2×D
Hint #2
In eq.3, replace E with 2×D: D – B = B – D + 2×D which becomes D – B = B + D In the above equation, add B to each side, and subtract D from each side: D – B + B – D = B + D + B – D which becomes 0 = 2×B which means 0 = B
Hint #3
eq.5 may be written as: 10×D + E = (D × E) + (E × F) In the equation above, substitute 2×D for E: 10×D + 2×D = (D × 2×D) + (2×D × F) which becomes eq.5a) 12×D = 2×D × (D + F)
Hint #4
Since D ≠ 0 (from eq.4), divide both sides of eq.5a by 2×D: 12×D ÷ 2×D = 2×D × (D + F) ÷ 2×D which becomes 6 = D + F Subtract D from both sides of the above equation: 6 – D = D + F – D which makes eq.5b) 6 – D = F
Hint #5
Substitute (6 – D) for F (from eq.5b), and 0 for B in eq.4: (6 – D) ÷ D = A + 0 + C which becomes eq.4a) (6 – D) ÷ D = A + C
Hint #6
Substitute (6 – D) ÷ D for A + C (from eq.4a), and 0 for B in eq.1: (6 – D) ÷ D = 0 + D which becomes (6 – D) ÷ D = D Multiply both sides of the equation above by D: D × (6 – D) ÷ D = D × D which becomes 6 – D = D² Add D to both sides, and subtract 6 from both sides: 6 – D + D – 6 = D² + D – 6 which becomes eq.1a) 0 = D² + D – 6
Hint #7
eq.1a is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for D in eq.1a yields: D = { (–1)×(1) ± sq.rt.[(1)² – (4 × (1) × (–6))] } ÷ (2 × (1)) which becomes D = {–1 ± sq.rt.(1 – (–24)} ÷ 2 which becomes D = {–1 ± sq.rt.(25)} ÷ 2 which becomes D = (–1 ± 5) ÷ 2 In the above equation, either D = (–1 + 5) ÷ 2 = 4 ÷ 2 = 2 or D = (–1 – 5) ÷ 2 = –6 ÷ 2 = –3 Since D must be non-negative, then D ≠ –3 and, therefore D = 2 making E = 2×D = 2 × 2 = 4
Hint #8
Substitute 2 for D in eq.5b: 6 – 2 = F which makes 4 = F
Hint #9
According to the definition of logarithms, eq.6 may be re-written as: E ^ A = F ("E ^ A" means "E raised to the power of A") Substitute 4 for both E and F in the above equation: 4 ^ A = 4 which means A = 1
Solution
Substitute 1 for A, 0 for B, and 2 for D in eq.1: 1 + C = 0 + 2 which becomes 1 + C = 2 Subtract 1 from both sides of the above equation: 1 + C – 1 = 2 – 1 which makes C = 1 and makes ABCDEF = 101244