Puzzle for July 22, 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
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Hint #1
In eq.4, replace B + C with F (from eq.2): E + F = A + F Subtract F from each side of the equation above: E + F - F = A + F - F which makes E = A
Hint #2
In eq.3, replace E with A: C = A + A which makes C = 2×A
Hint #3
eq.1 may be written as: D + A + B + C + E + F = 16 In the above equation, substitute D for A + B + C + E + F (from eq.5): D + D = 16 which makes 2×D = 16 Divide both sides by 2: 2×D ÷ 2 = 16 ÷ 2 which makes D = 8
Hint #4
eq.5 may be written: D = A + E + B + C + F Substitute 8 for D, C for A + E (from eq.3), and B + C for F (from eq.2) into the above equation: 8 = C + B + C + B + C which becomes eq.5a) 8 = 3×C + 2×B
Hint #5
Substitute (2×A) for C in eq.5a: 8 = 3×(2×A) + 2×B which becomes 8 = 6×A + 2×B Subtract 6×A from each side of the equation above: 8 - 6×A = 6×A + 2×B - 6×A which becomes 8 - 6×A = 2×B Divide both sides by 2: (8 - 6×A) ÷ 2 = 2×B ÷ 2 which becomes eq.5b) 4 - 3×A = B
Hint #6
Substitute 4 - 3×A for B (from eq.5b), and 2×A for C in eq.2: F = 4 - 3×A + 2×A which becomes eq.2a) F = 4 - A
Hint #7
Substitute 4 - 3×A for B (from eq.5b), 4 - A for F (from eq.2a), 8 for D, and 2×A for C in eq.6: 4 - 3×A + 4 - A = 8 ÷ 2×A which becomes 8 - 4×A = 8 ÷ 2×A Multiply both sides of the above equation by 2×A: 2×A × (8 - 4×A) = 2×A × (8 ÷ 2×A) which becomes 16×A - 8×A² = 8 Subtract 8 from each side: 16×A - 8×A² - 8 = 8 - 8 which becomes 16×A - 8×A² - 8 = 0 which may be written as -8×A² + 16×A - 8 = 0 Divide both sides by (-8): (-8×A² + 16×A - 8) ÷ (-8) = 0 ÷ (-8) which becomes eq.6a) A² - 2×A + 1 = 0
Hint #8
eq.6a is a quadratic equation in standard form. The quadratic equation solution formula could be used to solve for A in eq.6a. However, eq.6a can easily be factored into the product of two expressions: (A - 1) × (A - 1) = 0 The above equation makes: (A - 1) = 0 which means A = 1
Solution
Since A = 1, then: B = 4 - 3×1 = 4 - 3 = 1 (from eq.5b) C = 2×A = 2×1 = 2 E = A = 1 F = 4 - A = 4 - 1 = 3 (from eq.2a) and ABCDEF = 112813