Puzzle for December 11, 2022 ( )
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
Help Area
Hint #1
eq.4 may be written as: A + E = C + F + D In the above equation, replace C + F with A + B (from eq.3): A + E = A + B + D Subtract A from each side of the equation above: A + E - A = A + B + D - A which becomes eq.4a) E = B + D
Hint #2
In eq.2, replace E with B + D (from eq.4a): B + D - A = B - D Subtract B from both sides of the above equation: B + D - A - B = B - D - B which becomes D - A = -D Add A and D to both sides: D - A + A + D = -D + A + D which makes eq.2a) 2×D = A
Hint #3
In eq.5, replace A + B with C + F (from eq.3): C - E = C + F - C which becomes C - E = F Add E to both sides of the equation above: C - E + E = F + E which becomes eq.5a) C = F + E
Hint #4
In eq.4, substitute 2×D for A, and F + E for C (from eq.5a): 2×D + E = F + E + D + F which becomes 2×D + E = 2×F + E + D Subtract E and D from each side of the above equation: 2×D + E - E - D = 2×F + E + D - E - D which makes D = 2×F
Hint #5
Substitute (2×F) for D in eq.2a: 2×(2×F) = A which makes 4×F = A
Hint #6
Add A and D to both sides of eq.2: E - A + A + D = B - D + A + D which becomes E + D = B + A which may be written as eq.2b) D + E = A + B
Hint #7
eq.1 may be re-written as: A + B + D + E + C + F = 24 In the above equation, substitute A + B for D + E (from eq.2b), and for C + F (from eq.3): A + B + A + B + A + B = 24 which becomes 3×(A + B) = 24 Divide both sides by 3: 3×(A + B) ÷ 3 = 24 ÷ 3 which makes eq.1a) A + B = 8
Hint #8
Substitute 8 for A + B (from eq.1a) into eq.3: 8 = C + F Subtract F from both sides of the above equation: 8 - F = C + F - F which becomes eq.3a) 8 - F = C
Hint #9
Substitute 2×F for D, and 8 for A + B (from eq.1a) into eq.2b: 2×F + E = 8 Subtract 2×F from each side of the equation above: 2×F + E - 2×F = 8 - 2×F which becomes eq.2c) E = 8 - 2×F
Hint #10
In eq.6, substitute (8 - F) for C (from eq.3a), 2×F for D, 8 for A + B (from eq.1a), and 8 - 2×F for E (from eq.2c): (8 - F) × 2×F = 8 + 8 - 2×F which becomes 16×F - 2×F² = 16 - 2×F which is the same as F×(16 - 2×F) = 16 - 2×F Divide both sides of the above equation by 16 - 2×F (assumes F ≠ 8): F×(16 - 2×F) ÷ (16 - 2×F) = (16 - 2×F) ÷ (16 - 2×F) which makes F = 1
Hint #11
Confirm: F ≠ 8 ... Substituting 8 for F in eq.2c would yield: E = 8 - 2×8 which would become E = 8 - 16 which would make E = -8 Since E must be non-negative, then: E ≠ -8 and therefore confirms: F ≠ 8 which means F = 1
Hint #12
Since F = 1, then: A = 4×F = 4 × 1 = 4 C = 8 - F = 8 - 1 = 7 (from eq.3a) D = 2×F = 2 × 1 = 2 E = 8 - 2×F = 8 - 2×1 = 8 - 2 = 6 (from eq.2c)
Solution
Substitute 4 for A in eq.1a: 4 + B = 8 Subtract 4 from both sides of the equation above: 4 + B - 4 = 8 - 4 which makes B = 4 and makes ABCDEF = 447261