Puzzle for March 5, 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.
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Hint #1
Add F to both sides of eq.2: C - F + F = A + F + F which becomes eq.2a) C = A + 2×F
Hint #2
In eq.3, replace C with A + 2×F (from eq.2a): B - A = A + 2×F - B Add A and B to both sides of the above equation: B - A + A + B = A + 2×F - B + A + B which becomes 2×B = 2×A + 2×F Divide both sides by 2: 2×B ÷ 2 = (2×A + 2×F) ÷ 2 which becomes eq.3a) B = A + F
Hint #3
In eq.4, substitute (A + F) for B (from eq.3a): D - A = A - (A + F) which becomes D - A = A - A - F which becomes eq.4a) D - A = -F
Hint #4
eq.5 may be written as: A + F = D - A + E In the above equation, replace D - A with -F (from eq.4a): A + F = -F + E Add F to both sides: A + F + F = -F + E + F which becomes eq.5a) A + 2×F = E
Hint #5
In eq.2a, substitute E for A + 2×F (from eq.5a): C = E
Hint #6
Add A and F to both sides of eq.4a: D - A + A + F = -F + A + F which becomes eq.4b) D + F = A Substitute D + F for A (from eq.4b) into eq.5a: D + F + 2×F = E which makes D + 3×F = E and also makes eq.5b) D + 3×F = E = C
Hint #7
Substitute D + F for A (from eq.4b) into eq.3a: B = D + F + F which becomes eq.3b) B = D + 2×F
Hint #8
Substitute D + F for A (from eq.4b), D + 2×F for B (from eq.3b), and D + 3×F for C and E (from eq.5b) in eq.1: D + F + D + 2×F + D + 3×F + D + D + 3×F + F = 35 which simplifies to 5×D + 10×F = 35 Divide both sides of the above equation by 5: (5×D + 10×F) ÷ 5 = 35 ÷ 5 which becomes eq.1a) D + 2×F = 7
Hint #9
eq.6 may be written as: D × F = (A + B) ÷ 2 In the above equation, substitute D + F for A (from eq.4b), and D + 2×F for B (from eq.3b): D × F = (D + F + D + 2×F) ÷ 2 which becomes D × F = (2×D + 3×F) ÷ 2 Multiply both sides of the above equation by 2: 2 × (D × F) = 2 × (2×D + 3×F) ÷ 2 which becomes eq.6a) 2 × (D × F) = 2×D + 3×F
Hint #10
Subtract 2×F from both sides of eq.1a: D + 2×F - 2×F = 7 - 2×F which becomes eq.1b) D = 7 - 2×F
Hint #11
Substitute (7 - 2×F) for D (from eq.1b) in eq.6a: 2 × ((7 - 2×F) × F) = 2×(7 - 2×F) + 3×F which becomes 2 × (7×F - 2×F²) = 14 - 4×F + 3×F which becomes 14×F - 4×F² = 14 - F Subtract 14 from both sides, and add F to both sides: 14×F - 4×F² - 14 + F = 14 - F - 14 + F which becomes 15×F - 4×F² - 14 = 0 which may be written as eq.6b) -4×F² + 15×F - 14 = 0
Hint #12
eq.6b is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for F in eq.6b yields: F = { (-1)×(15) ± sq.rt.[(15)² - (4 × (-4) × (-14))] } ÷ (2 × (-4)) which becomes F = {-15 ± sq.rt.(225 - 224)} ÷ (-8) which becomes F = {-15 ± sq.rt.(1)} ÷ (-8) which becomes F = (-15 ± 1) ÷ (-8) In the above equation, either F = (-15 + 1) ÷ (-8) = (-14) ÷ (-8) = 1¾ or F = (-15 - 1) ÷ (-8) = (-16) ÷ (-8) = 2 Since F must be an integer, then F ≠ 1¾ and therefore makes F = 2
Hint #13
Substitute 2 for F in eq.1b: D = 7 - 2×2 which becomes D = 7 - 4 which makes D = 3
Solution
Since D = 3, and F = 2, then: A = D + F = 3 + 2 = 5 (from eq.4b) B = D + 2×F = 3 + 2×2 = 3 + 4 = 7 (from eq.3b) C = E = D + 3×F = 3 + 3×2 = 3 + 6 = 9 (from eq.5b) and ABCDEF = 579392