Puzzle for February 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.
* BC and EF are 2-digit numbers (not B×C or E×F).
Scratchpad
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Hint #1
In eq.4, substitute (D + F) for B (from eq.1): F - D = (D + F) ÷ D Multiply both sides of the above equation by D: D × (F - D) = D × (D + F) ÷ D which becomes eq.4a) (D × F) - D² = D + F
Hint #2
In eq.4a, subtract F from both sides, and add D² to both sides: (D × F) - D² - F + D² = D + F - F + D² which becomes (D × F) - F = D + D² which may be written as F × (D - 1) = D² + D Divide both sides by (D - 1): F × (D - 1) ÷ (D - 1) = (D² + D) ÷ (D - 1) which becomes eq.4b) F = (D² + D) ÷ (D - 1) (assumes D ≠ 1)
Hint #3
Begin confirming: D ≠ 1 ... Replacing D with 1 in eq.1 would yield: eq.1a) B = 1 + F Replacing D with 1 in eq.4 would yield: F - 1 = B ÷ 1 which would become eq.4c) F - 1 = B
Hint #4
Finish confirming: D ≠ 1 ... In eq.4c, replacing B with 1 + F (from eq.1a) would yield: F - 1 = 1 + F Subtracting F from each side of the equation above would yield: F - 1 - F = 1 + F - F which would make -1 = 1 Since -1 ≠ 1, then: D ≠ 1
Hint #5
Subtract D from both sides of eq.1: B - D = D + F - D which becomes eq.1b) B - D = F
Hint #6
In eq.6, replace B - D with F (from eq.1b): F = (E + F) ÷ D Multiply both sides of the above equation by D: D × F = D × (E + F) ÷ D which becomes D × F = E + F Subtract F from both sides: D × F - F = E + F - F which becomes eq.6a) F × (D - 1) = E
Hint #7
eq.3 may be written as: 10×B + C - (10×E + F) = B + E which is equivalent to 10×B + C - 10×E - F = B + E In the above equation, subtract B from both sides, and add 10×E and F to both sides: 10×B + C - 10×E - F - B + 10×E + F = B + E - B + 10×E + F which becomes eq.3a) 9×B + C = 11×E + F
Hint #8
In eq.3a, substitute D + F for B (from eq.1), and E - F for C (from eq.2): 9×(D + F) + E - F = 11×E + F which becomes 9×D + 9×F + E - F = 11×E + F which becomes 9×D + 8×F + E = 11×E + F Subtract E and F from both sides of the equation above: 9×D + 8×F + E - E - F = 11×E + F - E - F which becomes eq.3b) 9×D + 7×F = 10×E
Hint #9
Substitute (F × (D - 1)) for E (from eq.6a) into eq.3b: 9×D + 7×F = 10×(F × (D - 1)) which becomes 9×D + 7×F = 10×F×D - 10×F Subtract 7×F from each side of the above equation: 9×D + 7×F - 7×F = 10×F×D - 10×F - 7×F which becomes 9×D = 10×F×D - 17×F which may be written as 9×D = F×(10×D - 17) Since (10×D - 17) ≠ 0 (because D must be an integer), divide both sides by (10×D - 17): 9×D ÷ (10×D - 17) = F×(10×D - 17) ÷ (10×D - 17) which becomes eq.3c) 9×D ÷ (10×D - 17) = F
Hint #10
Substitute 9×D ÷ (10×D - 17) for F (from eq.3c) into eq.4b: 9×D ÷ (10×D - 17) = (D² + D) ÷ (D - 1) Cross-multiply both sides of the above equation by (D - 1) × (10×D - 17): 9×D ÷ (10×D - 17) × (D - 1) × (10×D - 17) = (D² + D) ÷ (D - 1) × (D - 1) × (10×D - 17) which becomes 9×D × (D - 1) = (D² + D) × (10×D - 17) which becomes 9×D² - 9×D = 10×D³ + 10×D² - 17×D² - 17×D which becomes eq.4d) 9×D² - 9×D = 10×D³ - 7×D² - 17×D
Hint #11
In eq.4d, subtract 9×D² from both sides, and add 9×D to both sides: 9×D² - 9×D - 9×D² + 9×D = 10×D³ - 7×D² - 17×D - 9×D² + 9×D which becomes 0 = 10×D³ - 16×D² - 8×D Since D ≠ 0 (from eq.4, eq.5, and eq.6), divide both sides by 2×D: 0 ÷ 2×D = (10×D³ - 16×D² - 8×D) ÷ 2×D which becomes eq.4e) 0 = 5×D² - 8×D - 4
Hint #12
eq.4e is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for D in eq.4e yields: D = { (-1)×(-8) ± sq.rt.[(-8)² - (4 × 5 × (-4))] } ÷ (2 × 5) which becomes D = {8 ± sq.rt.(64 - (-80))} ÷ 10 which becomes D = {8 ± sq.rt.(144)} ÷ 10 which becomes D = (8 ± 12) ÷ 10 In the above equation, either D = (8 + 12) ÷ 10 = 20 ÷ 10 = 2 or D = (8 - 12) ÷ 10 = -4 ÷ 10 = -0.4 Since D must be a non-negative integer, then D ≠ -0.4 and therefore makes D = 2
Hint #13
Substitute 2 for D in eq.4b: F = (2² + 2) ÷ (2 - 1) which becomes F = (4 + 2) ÷ 1 which becomes F = 6 ÷ 1 which makes F = 6
Hint #14
Substitute 6 for F, and 2 for D in eq.6a: 6 × (2 - 1) = E which becomes 6 × (1) = E which makes 6 = E
Hint #15
Substitute 2 for D, and 6 for F in eq.1b: B - 2 = 6 Add 2 to both sides of the equation above: B - 2 + 2 = 6 + 2 which makes B = 8
Hint #16
Substitute 6 for both E and F in eq.2: C = 6 - 6 which means C = 0
Solution
Substitute 2 for D, 8 for B, and 6 for E in eq.5: A - 2 = (8 + 6) ÷ 2 which becomes A - 2 = (14) ÷ 2 which becomes A - 2 = 7 Add 2 to both sides of the above equation: A - 2 + 2 = 7 + 2 which makes A = 9 and makes ABCDEF = 980266