Puzzle for October 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.
* CD and AB are 2-digit numbers (not C×D or A×B).
Scratchpad
Help Area
Hint #1
In eq.4, replace E with B + F (from eq.2): B - F = B + F - B which becomes B - F = F Add F to both sides of the above equation: B - F + F = F + F which makes B = 2×F
Hint #2
In eq.2, replace B with 2×F: E = 2×F + F which makes E = 3×F
Hint #3
In eq.3, substitute 3×F for E: 3×F = A + C Subtract C from each side of the equation above: 3×F - C = A + C - C which becomes eq.3a) 3×F - C = A
Hint #4
Substitute (3×F - C) for A (from eq.3a) in eq.5: C - (3×F - C) = D - F which becomes C - 3×F + C = D - F which becomes 2×C - 3×F = D - F Add F to both sides of the above equation: 2×C - 3×F + F = D - F + F which becomes eq.5a) 2×C - 2×F = D
Hint #5
eq.6 may be written as: 10×C + D - (10×A + B) = C + E - B which becomes 10×C + D - 10×A - B = C + E - B In the above equation, add B to both sides, and subtract C from both sides: 10×C + D - 10×A - B + B - C = C + E - B + B - C which becomes eq.6a) 9×C + D - 10×A = E
Hint #6
Substitute 2×C - 2×F for D (from eq.5a), (3×F - C) for A (from eq.3a), and 3×F for E in eq.6a: 9×C + 2×C - 2×F - 10×(3×F - C) = 3×F which becomes 11×C - 2×F - 30×F + 10×C = 3×F which becomes 21×C - 32×F = 3×F Add 32×F to both sides of the above equation: 21×C - 32×F + 32×F = 3×F + 32×F which makes 21×C = 35×F Divide both sides by 21: 21×C ÷ 21 = 35×F ÷ 21 which makes C = 1⅔×F
Hint #7
Substitute 1⅔×F for C in eq.3a: 3×F - 1⅔×F = A which makes 1⅓×F = A
Hint #8
Substitute (1⅔×F) for C in eq.5a: 2×(1⅔×F) - 2×F = D which becomes 3⅓×F - 2×F = D which makes 1⅓×F = D
Solution
Substitute 1⅓×F for A and D, 2×F for B, 1⅔×F for C, and 3×F for E in eq.1: 1⅓×F + 2×F + 1⅔×F + 1⅓×F + 3×F + F = 31 which simplifies to 10⅓×F = 31 Divide both sides of the above equation by 10⅓: 10⅓×F ÷ 10⅓ = 31 ÷ 10⅓ which means F = 3 making A = D = 1⅓×F = 1⅓ × 3 = 4 B = 2×F = 2 × 3 = 6 C = 1⅔×F = 1⅔ × 3 = 5 E = 3×F = 3 × 3 = 9 and ABCDEF = 465493