Puzzle for May 7, 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.
* EF is a 2-digit number (not E×F).
Scratchpad
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Hint #1
Add C and D to both sides of eq.3: F - C + C + D = A - D + C + D which becomes eq.3a) F + D = A + C Add C to both sides of eq.5: B - C + D + C = A - D + E + F + C which becomes B + D = A - D + E + F + C which may be written as eq.5a) B + D = A + C - D + E + F
Hint #2
In eq.5a, replace A + C with F + D (from eq.3a): B + D = F + D - D + E + F which becomes eq.5b) B + D = 2×F + E
Hint #3
Add F to both sides of eq.4: E + F - D + F = B + D - F + F which becomes E + 2×F - D = B + D which may be written as eq.4a) 2×F + E - D = B + D
Hint #4
In eq.4a, replace 2×F + E with B + D (from eq.5b): B + D - D = B + D which becomes B = B + D Subtract B from each side of the equation above: B - B = B + D - B which makes 0 = D
Hint #5
In eq.2, substitute 0 for D: A - E = E - 0 which becomes A - E = E Add E to both sides of the above equation: A - E + E = E + E which makes eq.2a) A = 2×E
Hint #6
In eq.5, substitute 0 for D: B - C + 0 = A - 0 + E + F which becomes B - C = A + E + F Add C to both sides of the above equation: B - C + C = A + E + F + C which becomes B = A + E + F + C which is the same as eq.5c) B = A + C + E + F
Hint #7
eq.1 may be written as: B + A + C + E + F + D = 18 In the equation above, substitute B for A + C + E + F (from eq.5c), and 0 for D: B + B + 0 = 18 which becomes 2×B = 18 Divide both sides by 2: 2×B ÷ 2 = 18 ÷ 2 which makes B = 9
Hint #8
Substitute 9 for B, and 0 for D in eq.5b: 9 + 0 = 2×F + E which becomes 9 = 2×F + E Subtract 2×F from each side of the equation above: 9 - 2×F = 2×F + E - 2×F which becomes eq.5d) 9 - 2×F = E
Hint #9
Substitute 9 - 2×F for E (from eq.5d) in eq.2a: A = 2×(9 - 2×F) which becomes eq.2b) A = 18 - 4×F
Hint #10
Substitute 18 - 4×F for A (from eq.2b), and 0 for D in eq.3: F - C = 18 - 4×F - 0 which becomes F - C = 18 - 4×F In the equation above, add C and 4×F to both sides, and subtract 18 from both sides: F - C + C + 4×F - 18 = 18 - 4×F + C + 4×F - 18 which becomes eq.3b) 5×F - 18 = C
Hint #11
eq.6 may be written as: 10×E + F = (B - A) × C Substitute 9 - 2×F for E (from eq.5c), 9 for B, (18 - 4×F) for A (from eq.2b), and (5×F - 18) for C (from eq.3b) in the above equation: 10×(9 - 2×F) + F = (9 - (18 - 4×F)) × (5×F - 18) which becomes 90 - 20×F + F = (-9 + 4×F) × (5×F - 18) which becomes 90 - 19×F = -45×F + 162 + 20×F² - 72×F which becomes eq.6a) 90 - 19×F = -117×F + 162 + 20×F²
Hint #12
In eq.6a, subtract 90 from both sides, and add 19×F to both sides: 90 - 19×F - 90 + 19×F = -117×F + 162 + 20×F² - 90 + 19×F which becomes 0 = -98×F + 72 + 20×F² which may be written as 0 = 20×F² - 98×F + 72 Divide both sides of the above equation by 2: 0 = (20×F² - 98×F + 72) ÷ 2 which becomes eq.6b) 0 = 10×F² - 49×F + 36
Hint #13
eq.6b is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for F in eq.6b yields: F = { (-1)×(-49) ± sq.rt.[(-49)² - (4 × 10 × 36)] } ÷ (2 × 10) which becomes F = {49 ± sq.rt.(2401 - 1440)} ÷ 20 which becomes F = {49 ± sq.rt.(961)} ÷ 20 which becomes eq.6c) F = (49 ± 31) ÷ 20
Hint #14
In eq.6c, either: F = (49 + 31) ÷ 20 = 80 ÷ 20 = 4 or: F = (49 - 31) ÷ 20 = 18 ÷ 20 = 0.9 Since F must be an integer, then F ≠ 0.9 and therefore makes F = 4
Solution
Since F = 4, then: C = 5×F - 18 = 5×4 - 18 = 20 - 18 = 2 (from eq.3b) E = 9 - 2×F = 9 - 2×4 = 9 - 8 = 1 (from eq.5d) A = 2×E = 2 × 1 = 2 (from eq.2a) and ABCDEF = 292014