Puzzle for May 21, 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.
Once again, we extend our thanks to frequent contributor, Judah S (age 16), for sending us an interesting and fun puzzle! Thanks, Judah!
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Hint #1
Multiply both sides of eq.5 by F: E × F = (A ÷ F) × F which becomes eq.5a) E × F = A
Hint #2
Add A to both sides of eq.6: A + A = (F × F) - A + A which becomes 2×A = F × F In the above equation, substitute (E × F) for A (from eq.5a): 2×(E × F) = F × F which may be written as (2×E) × F = F × F Since F ≠ 0 (from eq.5), divide both sides by F: (2×E) × F ÷ F = F × F ÷ F which makes 2×E = F
Hint #3
In eq.5a, replace F with (2×E): E × (2×E) = A which becomes 2×E² = A
Hint #4
Multiply both sides of eq.4 by C: C × C = (D ÷ C) × C which becomes C² = D
Hint #5
In eq.3, substitute 2×E² for A, and C² for D: 2×E² + B = C + C² - E Subtract 2×E² from each side of the equation above: 2×E² + B - 2×E² = C + C² - E - 2×E² which becomes eq.3a) B = C + C² - E - 2×E²
Hint #6
In eq.2, substitute C² for D, and 2×E² for A: C² + E = 2×E² + C Subtract E and 2×E² from both sides of the above equation: C² + E - E - 2×E² = 2×E² + C - E - 2×E² which becomes eq.2a) C² - 2×E² = C - E
Hint #7
eq.3a may be written as: B = C - E + C² - 2×E² Substitute C - E for C² - 2×E² (from eq.2a) in the equation above: B = C - E + C - E which becomes eq.3b) B = 2×C - 2×E
Hint #8
Substitute C² for D, 2×C - 2×E for B (from eq.3b), and 2×E for F in eq.1: C² = 2×C - 2×E + C + 2×E which becomes C² = 3×C Since C ≠ 0 (from eq.4), divide both sides of the above equation by C: C² ÷ C = 3×C ÷ C which makes C = 3 and also makes D = C² = 3² = 9
Hint #9
Substitute 3 for C in eq.2a: 3² - 2×E² = 3 - E which becomes 9 - 2×E² = 3 - E In the equation above, subtract 9 from both sides, and add 2×E² to both sides: 9 - 2×E² - 9 + 2×E² = 3 - E - 9 + 2×E² which becomes 0 = -6 - E + 2×E² which may be written as eq.2b) 0 = 2×E² - E - 6
Hint #10
eq.2b is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for E in eq.2b yields: E = { (-1)×(-1) ± sq.rt.[(-1)² - (4 × 2 × (-6))] } ÷ (2 × 2) which becomes E = {1 ± sq.rt.(1 - (-48))} ÷ 4 which becomes E = {1 ± sq.rt.(49)} ÷ 4 which becomes eq.2c) E = (1 ± 7) ÷ 4
Hint #11
In eq.2c, either: E = (1 + 7) ÷ 4 = 8 ÷ 4 = 2 or E = (1 - 7) ÷ 4 = -6 ÷ 4 = -1½ Since E must be a non-negative integer, then: E ≠ -1½ Therefore: E = 2 making A = 2×E² = 2 × 2² = 2 × 4 = 8 F = 2×E = 2 × 2 = 4
Solution
Substitute 3 for C, and 2 for E in eq.3b: B = 2×3 - 2×2 which becomes B = 6 - 4 which makes B = 2 and makes ABCDEF = 823924