Puzzle for November 30, 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.3, substitute (A + F) for D (from eq.2): B - (A + F) = A - F which becomes B - A - F = A - F Add A and F to both sides of the above equation: B - A - F + A + F = A - F + A + F which makes B = 2×A
Hint #2
eq.6 may be re-written as: F = (A + B + D) ÷ 3 Multiply both sides of the above equation by 3: 3 × F = 3 × (A + B + D) ÷ 3 which becomes eq.6a) 3×F = A + B + D
Hint #3
In eq.6a, replace B with 2×A, and D with A + F (from eq.2): 3×F = A + 2×A + A + F which becomes 3×F = 4×A + F Subtract F from each side of the equation above: 3×F - F = 4×A + F - F which makes 2×F = 4×A Divide both sides by 2: 2×F ÷ 2 = 4×A ÷ 2 which makes F = 2×A
Hint #4
In eq.2, replace F with 2×A: D = A + 2×A which makes D = 3×A
Hint #5
In eq.5, substitute 3×A for D, and 2×A for F: C + E = 3×A + 2×A - C which becomes C + E = 5×A - C Add C to both sides of the equation above: C + E + C = 5×A - C + C which becomes eq.5a) 2×C + E = 5×A
Hint #6
Substitute 2×A for B in eq.4: 2×A + C = A + E Subtract A from each side of the above equation: 2×A + C - A = A + E - A which becomes eq.4a) A + C = E
Hint #7
Substitute A + C for E (from eq.4a) into eq.5a: 2×C + A + C = 5×A which becomes 3×C + A = 5×A Subtract A from both sides of the above equation: 3×C + A - A = 5×A - A which makes 3×C = 4×A Divide both sides of the above equation by 3: 3×C ÷ 3 = 4×A ÷ 3 which makes C = 1⅓×A
Hint #8
Substitute 1⅓×A for C in eq.4a: A + 1⅓×A = E which makes 2⅓×A = E
Solution
Substitute 2×A for B and F, 1⅓×A for C, 3×A for D, and 2⅓×A for E in eq.1: A + 2×A + 1⅓×A + 3×A + 2⅓×A + 2×A = 35 which simplifies to 11⅔×A = 35 Divide both sides of the above equation by 11⅔: 11⅔×A ÷ 11⅔ = 35 ÷ 11⅔ which means A = 3 making B = F = 2×A = 2 × 3 = 6 C = 1⅓×A = 1⅓ × 3 = 4 D = 3×A = 3 × 3 = 9 E = 2⅓×A = 2⅓ × 3 = 7 and ABCDEF = 364976