Puzzle for October 28, 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 positive integer.
Scratchpad
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Hint #1
eq.5 may be written as: A = (B + C + E + F) ÷ 4 Multiply both sides of the above equation by 4: 4 × A = 4 × (B + C + E + F) ÷ 4 which becomes eq.5a) 4×A = B + C + E + F
Hint #2
In eq.5a, replace B + C with E + F (from eq.3): 4×A = E + F + E + F which becomes 4×A = 2×E + 2×F Divide both sides of the above equation by 2: 4×A ÷ 2 = (2×E + 2×F) ÷ 2 which becomes eq.5b) 2×A = E + F
Hint #3
Add D and E to both sides of eq.4: F - D + D + E = D - E + D + E which becomes F + E = 2×D which may be written as eq.4a) E + F = 2×D
Hint #4
In eq.4a, replace E + F with 2×A (from eq.5b): 2×A = 2×D Divide both sides of the above equation by 2: 2×A ÷ 2 = 2×D ÷ 2 which makes A = D
Hint #5
In eq.2, substitute A for D: B + A = A + F Subtract A from each side of the above equation: B + A - A = A + F - A which makes B = F
Hint #6
Substitute B for F in eq.3: E + B = B + C Subtract B from both sides of the above equation: E + B - B = B + C - B which makes E = C
Hint #7
Substitute A for D, and C for E in eq.6: B × A = A × C × C which may be written as B × A = A × C² Divide both sides of the above equation by A: (B × A) ÷ A = (A × C²) ÷ A which makes B = C² and also makes eq.6a) B = F = C²
Hint #8
Substitute C for E, and C² for F in eq.5b: 2×A = C + C² Divide both sides of the above equation by 2: 2×A ÷ 2 = (C + C²) ÷ 2 which becomes A = ½×(C + C²) and also makes eq.5c) A = D = ½×(C + C²)
Hint #9
Substitute ½×(C + C²) for A and D (from eq.5c), C² for B and F, and C for E in eq.1: ½×(C + C²) + C² + C + ½×(C + C²) + C + C² = 18 which becomes C + C² + 2×C² + 2×C = 18 which becomes 3×C + 3×C² = 18 Divide both sides of the above equation by 3: (3×C + 3×C²) ÷ 3 = 18 ÷ 3 which becomes C + C² = 6 Subtract 6 from both sides: C + C² - 6 = 6 - 6 which becomes C + C² - 6 = 0 which may be written as eq.1a) C² + C - 6 = 0
Hint #10
eq.1a is a quadratic equation in standard form. The quadratic equation solution formula could be used to solve for C in eq.1a. However, eq.1a can easily be factored into the product of two expressions: C² + C - 6 = 0 may be written as (C + 3) × (C - 2) = 0 The above equation makes either: (C + 3) = 0 which would make C = -3 or: (C - 2) = 0 which would make C = 2 Since C is non-negative, then: C ≠ -3 and therefore makes C = 2
Solution
Since C = 2, then: A = D = ½×(C + C²) = ½×(2 + 2²) = ½×(2 + 4) = ½×(6) = 3 (from eq.5c) B = F = C² = 2² = 4 (from eq.6a) E = C = 2 and ABCDEF = 342324