Puzzle for October 17, 2021 ( )
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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.
* BC and EF are 2-digit numbers (not B×C or E×F).
** "C mod B" equals the remainder of C ÷ B.
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Hint #1
In eq.6, substitute (B × E) for C (from eq.5): eq.6a) D = (B × E) mod B Since (B × E) ÷ B = E then, remainder of (B × E) ÷ B = 0 which means (B × E) mod B = 0 In eq.6a, substitute 0 for (B × E) mod B: D = 0
Hint #2
In eq.2, replace D with 0: E = B + 0 which makes E = B
Hint #3
eq.4 may be written as: 10×B + C + D = A + 10×E + F In the equation above, substitute 0 for D, and B for E: 10×B + C + 0 = A + 10×B + F which becomes 10×B + C = A + 10×B + F Subtract 10×B from both sides: 10×B + C – 10×B = A + 10×B + F – 10×B which becomes eq.4a) C = A + F
Hint #4
Substitute 0 for D in eq.3: F – A + 0 = A + E which becomes F – A = A + E Add A to both sides of the above equation: F – A + A = A + E + A which becomes eq.3a) F = 2×A + E
Hint #5
Substitute 2×A + E for F (from eq.3a) in eq.4a: C = A + 2×A + E which becomes eq.4b) C = 3×A + E
Hint #6
Substitute E for B, 3×A + E for C (from eq.4b), 0 for D, and 2×A + E for F (from eq.3a) in eq.1: A + E + 3×A + E + 0 + E + 2×A + E = 24 which becomes 6×A + 4×E = 24 Divide both sides of the above equation by 2: (6×A + 4×E) ÷ 2 = 24 ÷ 2 which becomes eq.1a) 3×A + 2×E = 12
Hint #7
In eq.5, substitute E for B: C = E × E which may be written as eq.5a) C = E²
Hint #8
Substitute E² for C (from eq.5a) in eq.4b: E² = 3×A + E Subtract E from each side of the equation above: E² – E = 3×A + E – E which becomes eq.4c) E² – E = 3×A
Hint #9
Substitute E² – E for 3×A (from eq.4c) in eq.1a: E² – E + 2×E = 12 which becomes E² + E = 12 Subtract 12 from both sides of the equation above: E² + E – 12 = 12 – 12 which becomes eq.1b) E² + E – 12 = 0
Hint #10
eq.1b is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for E in eq.1b yields: E = { (–1)×(1) ± sq.rt.[(1)² – (4 × (1) × (–12))] } ÷ (2 × (1)) which becomes E = {–1 ± sq.rt.(1 – (–48))} ÷ 2 which becomes E = {–1 ± sq.rt.(49)} ÷ 2 which becomes E = (–1 ± 7) ÷ 2 In the above equation, either E = (–1 + 7) ÷ 2 = 6 ÷ 2 = 3 or E = (–1 – 7) ÷ 2 = –8 ÷ 2 = –4 Since E must be a non-negative integer, then E ≠ –4 and therefore makes E = 3 and also makes B = E = 3
Hint #11
Substitute 3 for E in eq.5a: C = 3² which makes C = 9
Hint #12
Substitute 3 for E in eq.4c: 3² – 3 = 3×A which becomes 9 – 3 = 3×A which makes 6 = 3×A Divide both sides of the above equation above by 3: 6 ÷ 3 = 3×A ÷ 3 which makes 2 = A
Solution
Substitute 2 for A, and 3 for E in eq.3a: F = 2×2 + 3 which becomes F = 4 + 3 which makes F = 7 and makes ABCDEF = 239037