Puzzle for December 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.
* CD and DE are 2-digit numbers (not C×D or D×E).
** "B mod C" equals the remainder of (B ÷ C).
Scratchpad
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Hint #1
Subtract C from both sides of eq.3: B + C – C = C × D – C which becomes B = C × D – C which may be written as B = C × (D – 1) Since C ≠ 0 (from eq.1), divide both sides of the above equation by C: B ÷ C = C × (D – 1) ÷ C which becomes eq.3a) B ÷ C = D – 1
Hint #2
In eq.1, replace B ÷ C with D – 1 (from eq.3a): A – D = D – 1 Add D to both sides of the above equation: A – D + D = D – 1 + D which becomes eq.1a) A = 2×D – 1
Hint #3
Since D – 1 is an integer, then from eq.3a: B ÷ C = an integer which means remainder (B ÷ C) = 0 which means B mod C = 0 In the above equation, replace B mod C with D – E (from eq.6): D – E = 0 Add E to both sides: D – E + E = 0 + E which makes D = E
Hint #4
In eq.2, substitute D for E: B ÷ F = D ÷ D which becomes B ÷ F = 1 Multiply both sides of the above equation by F: B ÷ F × F = 1 × F which makes B = F
Hint #5
Substitute D for E, and (2×D – 1) for A (from eq.1a) in eq.4: D + (D × F) = (2×D – 1) × D which may be written as D × (1 + F) = (2×D – 1) × D Since D ≠ 0 (from eq.2), divide both sides of the above equation by D: D × (1 + F) ÷ D = (2×D – 1) × D ÷ D which becomes 1 + F = 2×D – 1 Subtract 1 from both sides: 1 + F – 1 = 2×D – 1 – 1 which makes eq.4a) F = 2×D – 2 and also makes eq.4b) B = F = 2×D – 2
Hint #6
Substitute 2×D – 2 for B (from eq.4b) in eq.3: 2×D – 2 + C = C × D Subtract C from each side of the equation above: 2×D – 2 + C – C = C × D – C which becomes 2×D – 2 = C × D – C which may be written as 2 × (D – 1) = C × (D – 1) Divide both sides of the above equation by D – 1 (assumes D ≠ 1): 2 × (D – 1) ÷ (D – 1) = C × (D – 1) ÷ (D – 1) which makes 2 = C
Hint #7
Confirm: D ≠ 1 ... Substituting 1 for D in eq.4a would yield: F = 2×1 – 2 which would become F = 2 – 2 which would make F = 0 However, F ≠ 0 (from eq.2), therefore: D ≠ 1 which means C = 2
Hint #8
eq.5 may be written as: 10×C + D + 10×D + E = (A × B) – C + D + E which becomes 10×C + 11×D + E = (A × B) – C + D + E In the above equation, substitute 2 for C, and D for E: 10×2 + 11×D + D = (A × B) – 2 + D + D which becomes 20 + 12×D = (A × B) – 2 + 2×D Add 2 to both sides, and subtract 2×D from both sides: 20 + 12×D + 2 – 2×D = (A × B) – 2 + 2×D + 2 – 2×D which becomes eq.5a) 22 + 10×D = A × B
Hint #9
Substitute (2×D – 1) for A (from eq.1a), and (2×D – 2) for B (from eq.4b) in eq.5a: 22 + 10×D = (2×D – 1) × (2×D – 2) which becomes 22 + 10×D = 4×D² – 2×D – 4×D + 2 which becomes 22 + 10×D = 4×D² – 6×D + 2 Subtract 22 and 10×D from both sides of the above equation: 22 + 10×D – 22 – 10×D = 4×D² – 6×D + 2 – 22 – 10×D which becomes 0 = 4×D² – 16×D – 20 Divide both sides by 4: 0 ÷ 4 = (4×D² – 16×D – 20) ÷ 4 which becomes eq.5b) 0 = D² – 4×D – 5
Solution
eq.5b is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for D in eq.5b yields: D = { (–1)×(–4) ± sq.rt.[(–4)² – (4 × 1 × (–5))] } ÷ (2 × 1) which becomes D = {4 ± sq.rt.(16 – (–20))} ÷ 2 which becomes D = {4 ± sq.rt.(36)} ÷ 2 which becomes D = (4 ± 6) ÷ 2 In the above equation, either D = (4 + 6) ÷ 2 = 10 ÷ 2 = 5 or D = (4 – 6) ÷ 2 = –2 ÷ 2 = –1 Since D must be non-negative, then D ≠ –1 and therefore makes D = 5 making A = 2×D – 1 = 2×5 – 1 = 10 – 1 = 9 (from eq.1a) B = F = 2×D – 2 = 2×5 – 2 = 10 – 2 = 8 (from eq.4b) E = D = 5 and ABCDEF = 982558