Puzzle for December 4, 2022 ( )
Scratchpad
Find the 6-digit number ABCDEF by solving the following equations:
A, B, C, D, E, and F each represent a one-digit positive integer.
Once again, we thank Judah S (age 16) for contributing a fun and challenging puzzle. Thank you so much, Judah!
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Hint #1
Add D and A to both sides of eq.3: A - D + D + A = B - A + D + A which becomes eq.3a) 2×A = B + D Add F and A to both sides of eq.4: A - F + F + A = F - A - B + F + A which becomes eq.4a) 2×A = 2×F - B
Hint #2
In eq.4a, replace 2×A with B + D (from eq.3a): B + D = 2×F - B Add B to both sides of the above equation: B + D + B = 2×F - B + B which becomes eq.4b) 2×B + D = 2×F
Hint #3
Add the left and right sides of eq.2 to the left and right sides of eq.1: F + F = B + C + D + E which becomes eq.2a) 2×F = B + C + D + E
Hint #4
In eq.2a, replace 2×F with 2×B + D (from eq.4b): 2×B + D = B + C + D + E Subtract B and D from each side of the equation above: 2×B + D - B - D = B + C + D + E - B - D which becomes eq.2b) B = C + E
Hint #5
In eq.5, substitute C + E for B (from eq.2b): C + E - C = (C + E - E) × C which becomes E = C × C which may be written as E = C²
Hint #6
Substitute C² for E in eq.2b: eq.2c) B = C + C²
Hint #7
Substitute C + C² for B (from eq.2c) in eq.1: F = C + C² + C which becomes eq.1a) F = 2×C + C²
Hint #8
Substitute 2×C + C² for F (from eq.1a), and C² for E in eq.2: 2×C + C² = D + C² Subtract C² from each side of the above equation: 2×C + C² - C² = D + C² - C² which becomes 2×C = D
Hint #9
Substitute C + C² for B (from eq.2c), 2×C for D, C² for E, and (2×C + C²) for F (from eq.1a) in eq.6: C + C² + C = (2×C × C²) - (2×C + C²) which becomes 2×C + C² = 2×C³ - 2×C - C² Subtract 2×C and C² from both sides of the equation above: 2×C + C² - 2×C - C² = 2×C³ - 2×C - C² - 2×C - C² which becomes 0 = 2×C³ - 4×C - 2×C² Divide both sides by 2×C: 0 ÷ 2×C = (2×C³ - 4×C - 2×C²) ÷ 2×C which becomes 0 = C² - 2 - C which may be written as eq.6a) 0 = C² - C - 2
Hint #10
eq.6a is a quadratic equation in standard form. eq.6a may be written as the product of two factors: 0 = (C - 2) × (C + 1) To make the above equation true, either: C - 2 = 0 which makes C = 2 or C + 1 = 0 which makes C = -1 Since C must be positive, then C ≠ -1 and therefore makes C = 2
Hint #11
Since C = 2, then: B = C + C² = 2 + 2² = 2 + 4 = 6 (from eq.2c) D = 2×C = 2×2 = 4 E = C² = 2² = 4 F = 2×C + C² = 2×2 + 2² = 4 + 4 = 8 (from eq.1a)
Solution
Substitute 6 for B, and 4 for D in eq.3a: 2×A = 6 + 4 which becomes 2×A = 10 Divide both sides of the above equation by 2: 2×A ÷ 2 = 10 ÷ 2 which makes A = 5 and makes ABCDEF = 562448