Puzzle for October 15, 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 F to both sides of eq.3: E - F + F = A + F + F which becomes eq.3a) E = A + 2×F
Hint #2
In eq.4, replace E with A + 2×F (from eq.3a): A + A + 2×F = B - A + F which becomes 2×A + 2×F = B - A + F In the equation above, add A to both sides, and subtract F from both sides: 2×A + 2×F + A - F = B - A + F + A - F which becomes eq.4a) 3×A + F = B
Hint #3
In eq.5, substitute (A + 2×F) for E (from eq.3a), and C - F for D (from eq.2): A + (A + 2×F) - C = C + C - F - A - (A + 2×F) which becomes 2×A + 2×F - C = 2×C - F - A - A - 2×F which becomes 2×A + 2×F - C = 2×C - 3×F - 2×A Add C, 3×F, and 2×A to both sides of the above equation: 2×A + 2×F - C + C + 3×F + 2×A = 2×C - 3×F - 2×A + C + 3×F + 2×A which simplifies to 4×A + 5×F = 3×C Divide both sides by 3: (4×A + 5×F) ÷ 3 = 3×C ÷ 3 which becomes eq.5a) 1⅓×A + 1⅔×F = C
Hint #4
Substitute 1⅓×A + 1⅔×F for C (from eq.5a) in eq.2: D = 1⅓×A + 1⅔×F - F which becomes eq.2a) D = 1⅓×A + ⅔×F
Hint #5
Substitute 3×A + F for B (from eq.4a), 1⅓×A + 1⅔×F for C (from eq.5a), 1⅓×A + ⅔×F for D (from eq.2a), and A + 2×F for E (from eq.3a) in eq.1: A + 3×A + F + 1⅓×A + 1⅔×F + 1⅓×A + ⅔×F + A + 2×F + F = 23 which simplifies to eq.1a) 7⅔×A + 6⅓×F = 23
Hint #6
In eq.6, substitute (3×A + F) for B (from eq.4a), and A + 2×F for E (from eq.3a): (3×A + F) ÷ A = A + 2×F - F which becomes eq.6a) (3×A + F) ÷ A = A + F
Hint #7
Multiply both sides of eq.6a by A: A × (3×A + F) ÷ A = A × (A + F) which becomes 3×A + F = A² + A×F Subtract 3×A and A×F from each side of the equation above: 3×A + F - 3×A - A×F = A² + A×F - 3×A - A×F which becomes F - A×F = A² - 3×A which may be written as F × (1 - A) = A² - 3×A Divide both sides by (1 - A): F × (1 - A) ÷ (1 - A) = (A² - 3×A) ÷ (1 - A) which becomes eq.6b) F = (A² - 3×A) ÷ (1 - A) (assumes A ≠ 1)
Hint #8
Check: A ≠ 1 ... If A = 1, then substituting 1 for A in eq.6a would yield: (3×1 + F) ÷ 1 = 1 + F which would become 3 + F = 1 + F Subtracting F from each side of the above equation would yield: 3 + F - F = 1 + F - F which would make 3 = 1 Since 3 ≠ 1, then: A ≠ 1
Hint #9
Substitute ((A² - 3×A) ÷ (1 - A)) for F (from eq.6b) in eq.1a: 7⅔×A + 6⅓×((A² - 3×A) ÷ (1 - A)) = 23 Multiply both sides of the above equation by (1 - A): (1 - A) × (7⅔×A + 6⅓×((A² - 3×A) ÷ (1 - A))) = (1 - A) × 23 which becomes (1 - A) × 7⅔×A + (6⅓×A² - 19×A) = 23 - 23×A which becomes 7⅔×A - 7⅔×A² + 6⅓×A² - 19×A = 23 - 23×A which becomes eq.1b) -11⅓×A - 1⅓×A² = 23 - 23×A
Hint #10
Add 11⅓×A and 1⅓×A² to both sides of eq.1b: -11⅓×A - 1⅓×A² + 11⅓×A + 1⅓×A² = 23 - 23×A + 11⅓×A + 1⅓×A² which becomes 0 = 23 - 11⅔×A + 1⅓×A² which may be written as 0 = 1⅓×A² - 11⅔×A + 23 Multiply both sides of the above equation by 3 (to eliminate fractions, and make solving less difficult): 3 × 0 = 3 × (1⅓×A² - 11⅔×A + 23) which becomes eq.1c) 0 = 4×A² - 35×A + 69
Hint #11
eq.1c is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for A in eq.1c yields: A = { (-1)×(-35) ± sq.rt.[(-35)² - (4 × 4 × 69)] } ÷ (2 × 4) which becomes A = {35 ± sq.rt.(1225 - 1104)} ÷ 8 which becomes A = {35 ± sq.rt.(121)} ÷ 8 which becomes eq.1d) A = (35 ± 11) ÷ 8
Hint #12
In eq.1d, either: A = (35 + 11) ÷ 8 = 46 ÷ 8 = 5¾ or: A = (35 - 11) ÷ 8 = 24 ÷ 8 = 3 Since A must be an integer, then A ≠ 5¾ and therefore means A = 3
Hint #13
Substitute 3 for A in eq.6b: F = (3² - 3×3) ÷ (1 - 3) which becomes F = (9 - 9) ÷ (-2) which becomes F = 0 ÷ (-2) which means F = 0
Hint #14
Substitute 3 for A, and 0 for F in eq.4a: 3×3 + 0 = B which makes 9 = B
Hint #15
Substitute 3 for A, and 0 for F in eq.5a: 1⅓×3 + 1⅔×0 = C which becomes 4 + 0 = C which makes 4 = C
Hint #16
Substitute 3 for A, and 0 for F in eq.2a: D = 1⅓×3 + ⅔×0 which becomes D = 4 + 0 which makes D = 4
Solution
Substitute 3 for A, and 0 for F in eq.3a: E = 3 + 2×0 which becomes E = 3 + 0 which makes E = 3 and makes ABCDEF = 394430