Puzzle for May 15, 2022 ( )
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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.
We thank our frequent contributor Judah S (age 15) for sending us today's puzzle, and for every puzzle of the entire past week. Thank you so much, Judah!
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Hint #1
In eq.2, replace C with D – B (from eq.1): B + D = A – B + D – B which becomes B + D = A – 2×B + D In the above equation, subtract D from both sides, and add 2×B to both sides: B + D – D + 2×B = A – 2×B + D – D + 2×B which makes 3×B = A
Hint #2
In eq.4, replace A with 3×B: 3×B + B = (B × D) – 3×B which becomes 4×B = (B × D) – 3×B Add 3×B to both sides of the above equation: 4×B + 3×B = (B × D) – 3×B + 3×B which becomes 7×B = (B × D) Since B ≠ 0 (from eq.5), divide both sides of the above equation by B: 7×B ÷ B = (B × D) ÷ B which makes 7 = D
Hint #3
In eq.1, substitute 7 for D: eq.1a) C = 7 – B
Hint #4
Substitute (7 – B) for C, and 3×B for A in eq.6: B + (7 – B) = (3×B × B) – (7 – B) – E which becomes 7 = 3×B² – 7 + B – E In the above equation, subtract 7 from both sides, and add E to both sides: 7 – 7 + E = 3×B² – 7 + B – E – 7 + E which becomes eq.6a) E = 3×B² – 14 + B
Hint #5
Substitute 3×B for A, 7 for D, and 3×B² – 14 + B for E (from eq.6a) in eq.5: 3×B = (3×B + 7 + 3×B² + B – 14 – F) ÷ B which becomes 3×B² = 3×B + 7 + 3×B² + B – 14 – F which becomes 3×B² = 4×B – 7 + 3×B² – F In the above equation, subtract 3×B² from both sides, and add F to both sides: 3×B² – 3×B² + F = 4×B – 7 + 3×B² – F – 3×B² + F which becomes eq.5a) F = 4×B – 7
Hint #6
Substitute 3×B for A, (7 – B) for C (from eq.1a), 7 for D, and 4×B – 7 for F (from eq.5a) in eq.3: 3×B × 3×B = ((7 – B) × 7) + 4×B – 7 which becomes 9×B² = 49 – 7×B + 4×B – 7 which becomes 9×B² = 42 – 3×B In the above equation, subtract 42 from both sides, and add 3×B to both sides: 9×B² – 42 + 3×B = 42 – 3×B – 42 + 3×B which becomes 9×B² – 42 + 3×B = 0 which may be written as 9×B² + 3×B – 42 = 0 Divide both sides by 3: (9×B² + 3×B – 42) ÷ 3 = 0 ÷ 3 which becomes eq.3a) 3×B² + B – 14 = 0
Hint #7
eq.3a is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for B in eq.3a yields: B = {(–1)×(1) ± sq.rt.[(1)² – (4 × 3 × (–14))]} ÷ (2 × 3) which becomes B = {–1 ± sq.rt.(1 – (–168))} ÷ 6 which becomes B = {–1 ± sq.rt.(169)} ÷ 6 which becomes B = (–1 ± 13) ÷ 6 In the above equation, either B = (–1 + 13) ÷ 6 = 12 ÷ 6 = 2 or B = (–1 – 13) ÷ 6 = –14 ÷ 6 = –2⅓ Since B must be a non-negative integer, then B ≠ –2⅓ and therefore makes B = 2
Solution
Since B = 2, then: A = 3×B = 3 × 2 = 6 C = 7 – B = 7 – 2 = 5 (from eq.1a) E = 3×B² – 14 + B = 3×2² – 14 + 2 = 3×4 – 12 = 12 – 12 = 0 (from eq.6a) F = 4×B – 7 = 4×2 – 7 = 8 – 7 = 1 (from eq.5a) and ABCDEF = 625701