Puzzle for December 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.
Scratchpad
Help Area
Hint #1
In eq.4, add D to both sides, and subtract F from both sides: D - F + D - F = E + F - D + D - F which becomes eq.4a) 2×D - 2×F = E
Hint #2
In eq.3, substitute (2×D - 2×F) for E (from eq.4a): (2×D - 2×F) - D = A - (2×D - 2×F) which becomes D - 2×F = A - 2×D + 2×F In the above equation, add 2×D to both sides, and subtract 2×F from both sides: D - 2×F + 2×D - 2×F = A - 2×D + 2×F + 2×D - 2×F which becomes eq.3a) 3×D - 4×F = A
Hint #3
In eq.2, substitute 2×D - 2×F for E (from eq.4a): D + 2×D - 2×F = A + F which becomes 3×D - 2×F = A + F Subtract F from each side of the equation above: 3×D - 2×F - F = A + F - F which becomes eq.2a) 3×D - 3×F = A
Hint #4
In eq.3a, replace A with 3×D - 3×F (from eq.2a): 3×D - 4×F = 3×D - 3×F In the equation above, subtract 3×D from both sides, and add 4×F to both sides: 3×D - 4×F - 3×D + 4×F = 3×D - 3×F - 3×D + 4×F which makes 0 = F
Hint #5
Substitute 0 for F in eq.2a: 3×D - 3×0 = A which becomes 3×D - 0 = A which makes 3×D = A
Hint #6
Substitute 0 for F in eq.4a: 2×D - 2×0 = E which becomes 2×D - 0 = E which makes 2×D = E
Hint #7
Substitute 2×D for E, 3×D for A, and 0 for F in eq.6: 2×D ÷ D = 3×D - C - 0 which becomes 2 = 3×D - C In the above equation, subtract 2 from both sides, and add C to both sides: 2 - 2 + C = 3×D - C - 2 + C which makes eq.6a) C = 3×D - 2
Hint #8
Substitute 3×D for A, and 0 for F, and (3×D - 2) for C (from eq.6a) in eq.5: 3×D - B + 0 = B - D - (3×D - (3×D - 2)) which becomes 3×D - B = B - D - (3×D - 3×D + 2) which becomes 3×D - B = B - D - (0 + 2) which becomes 3×D - B = B - D - 2 Add B, D, and 2 to both sides of the above equation: 3×D - B + B + D + 2 = B - D - 2 + B + D + 2 which becomes 4×D + 2 = 2×B Divide both sides by 2: (4×D + 2) ÷ 2 = 2×B ÷ 2 which makes eq.5a) 2×D + 1 = B
Hint #9
Substitute 3×D for A, 2×D + 1 for B (from eq.5a), 3×D - 2 for C (from eq.6a), 2×D for E, and 0 for F in eq.1: 3×D + 2×D + 1 + 3×D - 2 + D + 2×D + 0 = 32 which simplifies to 11×D - 1 = 32 Add 1 to both sides of the above equation: 11×D - 1 + 1 = 32 + 1 which makes 11×D = 33 Divide both sides by 11: 11×D ÷ 11 = 33 ÷ 11 which means D = 3
Solution
Since D = 3, then: A = 3×D = 3 × 3 = 9 B = 2×D + 1 = 2×3 + 1 = 6 + 1 = 7 (from eq.5a) C = 3×D - 2 = 3×3 - 2 = 9 - 2 = 7 (from eq.6a) E = 2×D = 2 × 3 = 6 and ABCDEF = 977360