Puzzle for October 1, 2023 ( )
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 and B to both sides of eq.2: B - A + A + B = C - B + A + B which becomes eq.2a) 2×B = C + A Add B and D to both sides of eq.5: C + F - B + B + D = B - D + E + B + D which becomes eq.5a) C + F + D = 2×B + E
Hint #2
In eq.5a, replace 2×B with C + A (from eq.2a): C + F + D = C + A + E Subtract C and A from each side of the equation above: C + F + D - C - A = C + A + E - C - A which becomes eq.5b) F + D - A = E
Hint #3
Add D to both sides of eq.3: F - A + D = A - E + D which may be written as eq.3a) F + D - A = A - E + D
Hint #4
In eq.3a, replace F + D - A with E (from eq.5b): E = A - E + D Add E to both sides of the above equation: E + E = A - E + D + E which becomes eq.3b) 2×E = A + D
Hint #5
In eq.4, substitute 2×E for A + D (from eq.3b): E - D = 2×E - E which becomes E - D = E Subtract E from each side of the equation above: E - D - E = E - E which makes -D = 0 which means D = 0
Hint #6
Substitute 0 for D in eq.3b: 2×E = A + 0 which makes 2×E = A
Hint #7
Substitute 2×E for A in eq.3: F - 2×E = 2×E - E which becomes F - 2×E = E Add 2×E to both sides of the above equation: F - 2×E + 2×E = E + 2×E which makes F = 3×E
Hint #8
Substitute 2×E for A, and 0 for D in eq.6: B × E = 2×E + C + 0 which becomes eq.6a) B × E = 2×E + C
Hint #9
Substitute 2×E for A, 0 for D, and 3×E for F in eq.1: 2×E + B + C + 0 + E + 3×E = 23 which becomes 6×E + B + C = 23 which may be written as eq.1a) 4×E + B + 2×E + C = 23
Hint #10
Substitute (B × E) for 2×E + C (from eq.6a) in eq.1a: 4×E + B + (B × E) = 23 Subtract 4×E from both sides of the equation above: 4×E + B + (B × E) - 4×E = 23 - 4×E which becomes B + (B × E) = 23 - 4×E which may be written as B × (1 + E) = 23 - 4×E Divide both sides by (1 + E): B × (1 + E) ÷ (1 + E) = (23 - 4×E) ÷ (1 + E) which becomes eq.1b) B = (23 - 4×E) ÷ (1 + E)
Hint #11
Subtract 6×E and B from both sides of eq.1a: 4×E + B + 2×E + C - 6×E - B = 23 - 6×E - B which becomes eq.1c) C = 23 - 6×E - B
Hint #12
Substitute 23 - 6×E - B for C (from eq.1c), and 2×E for A in eq.2a: 2×B = 23 - 6×E - B + 2×E which becomes 2×B = 23 - 4×E - B Add B to both sides of the above equation: 2×B + B = 23 - 4×E - B + B which becomes eq.2b) 3×B = 23 - 4×E
Hint #13
Substitute ((23 - 4×E) ÷ (1 + E)) for B (from eq.1b) into eq.2b: 3×((23 - 4×E) ÷ (1 + E)) = 23 - 4×E which becomes (69 - 12×E) ÷ (1 + E) = 23 - 4×E Multiply both sides of the above equation by (1 + E): ((69 - 12×E) ÷ (1 + E)) × (1 + E) = (23 - 4×E) × (1 + E) which becomes 69 - 12×E = 23 + 23×E - 4×E - 4×E² which becomes 69 - 12×E = 23 + 19×E - 4×E² Subtract 69 from both sides, and add 12×E to both sides: 69 - 12×E - 69 + 12×E = 23 + 19×E - 4×E² - 69 + 12×E which becomes 0 = -46 + 31×E - 4×E² which may be written as eq.2c) 0 = -4×E² + 31×E - 46
Hint #14
eq.2c is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for E in eq.2c yields: E = { (-1)×31 ± sq.rt.[31² - (4 × (-4) × (-46))] } ÷ (2 × (-4)) which becomes E = {-31 ± sq.rt.(961 - 736)} ÷ (-8) which becomes E = {-31 ± sq.rt.(225)} ÷ (-8) which becomes eq.2d) E = (-31 ± 15) ÷ (-8)
Hint #15
In eq.2d, either: E = (-31 + 15) ÷ (-8) = -16 ÷ (-8) = 2 or: E = (-31 - 15) ÷ (-8) = -46 ÷ (-8) = 5¾ Since E must be an integer, then E ≠ 5¾ and therefore makes E = 2
Hint #16
Since E = 2, then: A = 2×E = 2 × 2 = 4 B = (23 - 4×E) ÷ (1 + E) = (23 - 4×2) ÷ (1 + 2) = (23 - 8) ÷ 3 = 15 ÷ 3 = 5 (from eq.1b) F = 3×E = 3 × 2 = 6
Solution
Substitute 2 for E, and 5 for B in eq.1c: C = 23 - 6×2 - 5 which becomes C = 18 - 12 which makes C = 6 and makes ABCDEF = 456026