Puzzle for November 12, 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
Add C and E to both sides of eq.3: B - C + C + E = A - E + C + E which becomes eq.3a) B + E = A + C
Hint #2
eq.6 may be written as: C = (B + D + E + F) ÷ 4 Multiply both sides of the above equation by 4: 4 × C = 4 × (B + D + E + F) ÷ 4 which becomes 4×C = B + D + E + F which may be written as eq.6a) 4×C = B + E + D + F
Hint #3
In eq.6a, replace B + E with A + C (from eq.3a): 4×C = A + C + D + F Subtract C from each side of the equation above: 4×C - C = A + C + D + F - C which becomes 3×C = A + D + F which may be written as eq.6b) A + D + F = 3×C
Hint #4
Add the left and right sides of eq.3a to the left and right sides of eq.6b, respectively: A + D + F + B + E = 3×C + A + C which becomes A + D + F + B + E = 4×C + A which may be written as eq.6c) A + B + F + D + E = 4×C + A
Hint #5
Add C and F to both sides of eq.5: A + B - C + C + F = C + D - F + C + F which becomes eq.5a) A + B + F = 2×C + D
Hint #6
In eq.6c, replace A + B + F with 2×C + D (from eq.5a): 2×C + D + D + E = 4×C + A which becomes 2×C + 2×D + E = 4×C + A Subtract 2×C from each side of the above equation: 2×C + 2×D + E - 2×C = 4×C + A - 2×C which becomes eq.6d) 2×D + E = 2×C + A
Hint #7
Add D and C to both sides of eq.2: E - D + D + C = D - C + D + C which becomes eq.2a) E + C = 2×D
Hint #8
In eq.6d, substitute E + C for 2×D (from eq.2a): E + C + E = 2×C + A which becomes 2×E + C = 2×C + A Subtract C from each side of the equation above: 2×E + C - C = 2×C + A - C which becomes 2×E = C + A which may be re-written as eq.6e) 2×E = A + C
Hint #9
Substitute 2×E for A + C (from eq.6e) into eq.3a: B + E = 2×E Subtract E from both sides of the above equation: B + E - E = 2×E - E which makes B = E
Hint #10
Add D and F to both sides of eq.4: D + F + D + F = B + C - D - F + D + F which becomes eq.4a) 2×D + 2×F = B + C
Hint #11
Substitute E + C for 2×D (from eq.2a), and E for B in eq.4a: E + C + 2×F = E + C Subtract E and C from each side of the equation above: E + C + 2×F - E - C = E + C - E - C which makes 2×F = 0 which means F = 0
Hint #12
Substitute 0 for F in eq.6b: A + D + 0 = 3×C which becomes A + D = 3×C Subtract A from both sides of the above equation: A + D - A = 3×C - A which becomes eq.6f) D = 3×C - A
Hint #13
Substitute (3×C - A) for D (from eq.6f) in eq.6d: 2×(3×C - A) + E = 2×C + A which becomes 6×C - 2×A + E = 2×C + A In the above equation, subtract 6×C from both sides, and add 2×A to both sides: 6×C - 2×A + E - 6×C + 2×A = 2×C + A - 6×C + 2×A which becomes eq.6g) E = 3×A - 4×C
Hint #14
Substitute (3×A - 4×C) for E (from eq.6g) in eq.6e: 2×(3×A - 4×C) = A + C which becomes 6×A - 8×C = A + C In the equation above, add 8×C to both sides, and subtract A from both sides: 6×A - 8×C + 8×C - A = A + C + 8×C - A which becomes 5×A = 9×C Divide both sides by 5: 5×A ÷ 5 = 9×C ÷ 5 which makes A = 1⅘×C
Hint #15
Substitute 1⅘×C for A in eq.6f: D = 3×C - 1⅘×C which makes D = 1⅕×C
Hint #16
Substitute (1⅘×C) for A in eq.6g: E = 3×(1⅘×C) - 4×C which becomes E = 5⅖×C - 4×C which makes E = 1⅖×C and also makes B = E = 1⅖×C
Solution
Substitute 1⅘×C for A, 1⅖×C for B and E, 1⅕×C for D, and 0 for F in eq.1: 1⅘×C + 1⅖×C + C + 1⅕×C + 1⅖×C + 0 = 34 which simplifies to 6⅘×C = 34 Divide both sides of the above equation by 6⅘: 6⅘×C ÷ 6⅘ = 34 ÷ 6⅘ which means C = 5 making A = 1⅘×C = 1⅘ × 5 = 9 B = E = 1⅖×C = 1⅖ × 5 = 7 D = 1⅕×C = 1⅕ × 5 = 6 and ABCDEF = 975670