Puzzle for November 19, 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.1, replace D + E + F with A + B + C (from eq.5): A + B + C + A + B + C = 32 which may be written as 2×(A + B + C) = 32 Divide both sides of the above equation by 2: 2×(A + B + C) ÷ 2 = 32 ÷ 2 which becomes eq.1a) A + B + C = 16
Hint #2
In eq.5, replace A + B + C with 16 (from eq.1a): eq.5a) D + E + F = 16
Hint #3
eq.1 may be written as: C + D + A + B + E + F = 32 In the above equation, substitute C + D for A + B + E + F (from eq.4): C + D + C + D = 32 which may be written as 2×(C + D) = 32 Divide both sides of the above equation by 2: 2×(C + D) ÷ 2 = 32 ÷ 2 which becomes eq.1b) C + D = 16
Hint #4
Substitute 16 for C + D (from eq.1b) in eq.4: eq.4a) 16 = A + B + E + F
Hint #5
Add D and A to both sides of eq.3: A - D + D + A = C - A + D + A which becomes 2×A = C + D Substitute 16 for C + D (from eq.1b) into the equation above: 2×A = 16 Divide both sides by 2: 2×A ÷ 2 = 16 ÷ 2 which makes A = 8
Hint #6
Substitute 8 for A, and F for B + E (from eq.2) in eq.4a: 16 = 8 + F + F which becomes 16 = 8 + 2×F Subtract 8 from each side of the above equation: 16 - 8 = 8 + 2×F - 8 which becomes 8 = 2×F Divide both sides by 2: 8 ÷ 2 = 2×F ÷ 2 which makes 4 = F
Hint #7
Substitute 4 for F in eq.5a: D + E + 4 = 16 Subtract D and 4 from both sides of the equation above: D + E + 4 - D - 4 = 16 - D - 4 which becomes eq.5b) E = 12 - D
Hint #8
Subtract D from each side of eq.1b: C + D - D = 16 - D which becomes eq.1c) C = 16 - D
Hint #9
Substitute (16 - D) for C (from eq.1c), (12 - D) for E (from eq.5b), 8 for A, and 4 for F in eq.6: (16 - D) × (12 - D) = 8 + D + 4 which becomes 192 - 16×D - 12×D + D² = 12 + D which becomes 192 - 28×D + D² = 12 + D Subtract 12 and D from both sides of the above equation: 192 - 28×D + D² - 12 - D = 12 + D - 12 - D which becomes 180 - 29×D + D² = 0 which may be written as eq.6a) D² - 29×D + 180 = 0
Hint #10
eq.6a is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for D in eq.6a yields: D = { (-1)×(-29) ± sq.rt.[(-29)² - (4 × 1 × 180)] } ÷ (2 × 1) which becomes D = {29 ± sq.rt.[841 - 720] } ÷ 2 which becomes D = {29 ± sq.rt.(121)} ÷ 2 which becomes eq.6b) D = (29 ± 11) ÷ 2
Hint #11
In eq.6b, either: D = (29 + 11) ÷ 2 = 40 ÷ 2 = 20 or D = (29 - 11) ÷ 2 = 18 ÷ 2 = 9 Since D must be a one-digit integer, then D ≠ 20 and therefore makes D = 9
Hint #12
Since D = 9, then: C = 16 - D = 16 - 9 = 7 (from eq.1c) E = 12 - D = 12 - 9 = 3 (from eq.5b)
Solution
Substitute 4 for F, and 3 for E in eq.2: 4 = B + 3 Subtract 3 from each side of the equation above: 4 - 3 = B + 3 - 3 which makes 1 = B and makes ABCDEF = 817934