Puzzle for November 29, 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
Add A to both sides of eq.2: F – A + A = D + E + A which is equivalent to F = E + A + D In the above equation, substitute (D × E) for A + D (from eq.4): F = E + (D × E) which may be written as eq.2a) F = (1 + D) × E
Hint #2
In eq.5, substitute (1 + D) × E for F (from eq.2a): (1 + D) × E ÷ E = C which becomes eq.5a) 1 + D = C
Hint #3
Add C and E to both sides of eq.3: C – D + C + E = F – C – E + C + E which becomes 2×C – D + E = F In the above equation, substitute (1 + D) for C (from eq.5a): 2×(1 + D) – D + E = F which is equivalent to 2×1 + 2×D – D + E = F which becomes eq.3a) 2 + D + E = F
Hint #4
Add A to both sides of eq.2: F – A + A = D + E + A which becomes F = D + E + A In the above equation, replace F with 2 + D + E (from eq.3a): 2 + D + E = D + E + A Subtract D and E from both sides of the above equation: 2 + D + E – D – E = D + E + A – D – E which simplifies to 2 = A
Hint #5
Substitute 2 for A in eq.1: D = 2 + B Subtract 2 from each side of the above equation: D – 2 = 2 + B – 2 which makes eq.1b) D – 2 = B Since B must be a non-negative integer, then in eq.1b: D – 2 = B ≥ 0 which means D – 2 ≥ 0 which makes eq.1c) D ≥ 2
Hint #6
Substitute 2 + D + E for F (from eq.3a) in eq.2a: 2 + D + E = (1 + D) × E which may be written as 2 + D + E = E + (E × D) Subtract D and E from both sides of the equation above: 2 + D + E – D – E = E + (E × D) – D – E which becomes eq.2b) 2 = (E × D) – D
Hint #7
Since D ≠ 0 (from eq.1c), divide both sides of eq.2b by D: 2 ÷ D = ((E × D) – D) ÷ D which becomes 2 ÷ D = E – 1 Add 1 to both sides of the equation above: (2 ÷ D) + 1 = E – 1 + 1 which becomes eq.2c) (2 ÷ D) + 1 = E
Hint #8
To make eq.2c true, check several possible values for D (D ≥ 2, from eq.1c) and for E: If D = 2, then E = (2 ÷ 2) + 1 = 1 + 1 = 2 If D = 3, then E = (2 ÷ 3) + 1 = ⅔ + 1 = 1.666666666667 If 3 ≤ D ≤ 9, then 1.666666666667 ≥ E ≥ 1.222222222222 Since E must be an integer, then E = 2 which means D = 2
Hint #9
Substitute 2 for D in eq.1b: 2 – 2 = B which makes 0 = B
Hint #10
Substitute 2 for D in eq.5a: 1 + 2 = C which makes 3 = C
Solution
Substitute 2 for D and E in eq.3a: 2 + 2 + 2 = F which makes 6 = F and makes ABCDEF = 203226