Puzzle for November 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.
Scratchpad
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Hint #1
In eq.3, replace C with A + F (from eq.2): A + F + D = B + F Subtract F from each side of the equation above: A + F + D - F = B + F - F which becomes eq.3a) A + D = B
Hint #2
In eq.6, replace B with A + D (from eq.3a): A + D - D = E ÷ A which becomes A = E ÷ A Multiply both sides of the above equation by A: A × A = A × (E ÷ A) which makes eq.6a) A² = E
Hint #3
In eq.1, substitute A + B + C for D + E + F (from eq.4): A + B + C + A + B + C = 22 which becomes 2×(A + B + C) = 22 Divide both sides of the above equation by 2: 2×(A + B + C) ÷ 2 = 22 ÷ 2 which becomes eq.1a) A + B + C = 11
Hint #4
Substitute D + E + F for A + B + C (from eq.4) in eq.1a: eq.1b) D + E + F = 11
Hint #5
Substitute A + D for B (from eq.3a), and A + F for C (from eq.2) into eq.4: D + E + F = A + A + D + A + F which becomes D + E + F = 3×A + D + F Subtract D and F from each side of the above equation: D + E + F - D - F = 2×A + B + F - D - F which makes E = 3×A
Hint #6
Substitute 3×A for E in eq.6a: A² = 3×A Since A ≠ 0 (from eq.6), divide both sides of the above equation by A: A² ÷ A = 3×A ÷ A which makes A = 3
Hint #7
Substitute 3 for A in eq.6a: 3² = E which makes 9 = E
Hint #8
Substitute 3 for A in eq.3a: eq.3b) 3 + D = B
Hint #9
Substitute 9 for E in eq.1b: D + 9 + F = 11 Subtract D and 9 from both sides of the equation above: D + 9 + F - D - 9 = 11 - D - 9 which becomes eq.1c) F = 2 - D
Hint #10
Substitute 3 for A, and 2 - D for F (from eq.1c) in eq.2: C = 3 + 2 - D which becomes eq.2a) C = 5 - D
Hint #11
Substitute 2 - D for F (from eq.1c), (3 + D) for B (from eq.3b), and (5 - D) for C (from eq.2a) in eq.5: 2 - D = (3 + D) ÷ (5 - D) Multiply both sides of the above equation by (5 - D): (2 - D) × (5 - D) = (3 + D) ÷ (5 - D) × (5 - D) which becomes 10 - 2×D - 5×D + D² = 3 + D which becomes 10 - 7×D + D² = 3 + D Subtract 3 and D from both sides of the equation above: 10 - 7×D + D² - 3 - D = 3 + D - 3 - D which becomes 7 - 8×D + D² = 0 which may be written as eq.5a) D² - 8×D + 7 = 0
Hint #12
eq.5a is a quadratic equation in standard form. The quadratic equation solution formula could be used to solve for D in eq.5a. However, eq.5a can be factored into the product of two expressions: D² - 8×D + 7 = 0 may be written as (D - 7) × (D - 1) = 0 The above equation makes either: (D - 7) = 0 which would make D = 7 or: (D - 1) = 0 which would make D = 1
Hint #13
Check: D = 7 ... Substituting 7 for D in eq.1c would yield: F = 2 - 7 which would make F = -5 Since F is non-negative, then: F ≠ -5 which means D ≠ 7 and therefore makes D = 1
Solution
Since D = 1, then: B = 3 + D = 3 + 1 = 4 (from eq.3b) C = 5 - D = 5 - 1 = 4 (from eq.2a) F = 2 - D = 2 - 1 = 1 (from eq.1c) and ABCDEF = 344191