Puzzle for June 11, 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
Add E and D to both sides of eq.2: D - E + E + D = A - D + E + D which becomes eq.2a) 2×D = A + E eq.5 may be written as: eq.5a) B + C + F = A + E + D
Hint #2
In eq.5a, replace A + E with 2×D (from eq.2a): B + C + F = 2×D + D which becomes eq.5b) B + C + F = 3×D
Hint #3
eq.1 may be written as: A + E + D + B + C + F = 24 In the above equation, replace A + E with 2×D (from eq.2a), and B + C + F with 3×D (from eq.5b): 2×D + D + 3×D = 24 which makes 6×D = 24 Divide both sides by 6: 6×D ÷ 6 = 24 ÷ 6 which makes D = 4
Hint #4
In eq.2a, substitute 4 for D: 2×4 = A + E which becomes 8 = A + E Subtract E from both sides of the equation above: 8 - E = A + E - E which becomes eq.2b) 8 - E = A
Hint #5
Substitute 8 - E for A (from eq.2b), and 4 for D in eq.3: E + F = 8 - E + 4 which becomes E + F = 12 - E Subtract E from both sides of the above equation: E + F - E = 12 - E - E which becomes eq.3a) F = 12 - 2×E
Hint #6
Substitute 8 - E for A (from eq.2b), 12 - 2×E for F (from eq.3a), and 4 for D in eq.4: 8 - E + 12 - 2×E - 4 = C + 4 which becomes 16 - 3×E = C + 4 Subtract 4 from each side of the above equation: 16 - 3×E - 4 = C + 4 - 4 which becomes eq.4a) 12 - 3×E = C
Hint #7
Substitute 12 - 3×E for C (from eq.4a), 12 - 2×E for F (from eq.3a), 8 - E for A (from eq.2b), and 4 for D in eq.5: B + 12 - 3×E + 12 - 2×E = 8 - E + 4 + E which becomes B + 24 - 5×E = 12 In the above equation, subtract 24 from both sides, and add 5×E to both sides: B + 24 - 5×E - 24 + 5×E = 12 - 24 + 5×E which becomes B = -12 + 5×E which is the same as eq.5c) B = 5×E - 12
Hint #8
Substitute 4 for D, (12 - 2×E) for F (from eq.3a), (8 - E) for A (from eq.2b), (5×E - 12) for B (from eq.5c), and (12 - 3×E) for C (from eq.4a) in eq.6: 4 × (12 - 2×E) = ((8 - E) × (5×E - 12)) + ((12 - 3×E) × E) which becomes 48 - 8×E = (40×E - 96 - 5×E² + 12×E) + (12×E - 3×E²) which becomes 48 - 8×E = 64×E - 96 - 8×E² In the equation above, subtract 48 from both sides, and add 8×E to both sides: 48 - 8×E - 48 + 8×E = 64×E - 96 - 8×E² - 48 + 8×E which becomes 0 = 72×E - 144 - 8×E² Divide both sides by (-8): 0 ÷ (-8) = (72×E - 144 - 8×E²) ÷ (-8) which becomes 0 = -9×E + 18 + E² which may be written as eq.6a) 0 = E² - 9×E + 18
Hint #9
eq.6a is a quadratic equation in standard form. The quadratic equation solution formula could be used to solve for E in eq.6a. However, the right side of eq.6a can be easily factored into the product of two expressions: 0 = (E - 3) × (E - 6) The above equation makes either: (E - 3) = 0 which makes E = 3 or: (E - 6) = 0 which makes E = 6
Hint #10
Check: E = 6 ... Substituting 6 for E in eq.4a would yield: 12 - 3×6 = C which would become 12 - 18 = C which would make -6 = C Since C is non-negative, then: C ≠ -6 which means E ≠ 6 and therefore makes E = 3
Solution
Since E = 3, then: A = 8 - E = 8 - 3 = 5 (from eq.2b) B = 5×E - 12 = 5×3 - 12 = 15 - 12 = 3 (from eq.5c) C = 12 - 3×E = 12 - 3×3 = 12 - 9 = 3 (from eq.4a) F = 12 - 2×E = 12 - 2×3 = 12 - 6 = 6 (from eq.3a) and ABCDEF = 533436