Puzzle for November 24, 2021 ( )
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.
Our thanks go out to Judah S (age 14) for this interesting puzzle. Thanks, Judah!
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Hint #1
In eq.2, replace F with A + D (from eq.1): C + D = A + A + D C + D = 2×A + D Subtract D from each side of the above equation: C + D – D = 2×A + D – D which makes C = 2×A
Hint #2
In eq.6, replace C with (2×A): A + (2×A) = (A × B) ÷ (2×A) which becomes 3×A = B ÷ 2 Multiply both sides of the above equation by 2: 2 × 3×A = 2 × (B ÷ 2) which makes 6×A = B
Hint #3
In eq.3, substitute 6×A for B: 6×A – A = A + E which becomes 5×A = A + E Subtract A from each side of the equation above: 5×A – A = A + E – A which makes 4×A = E
Hint #4
Substitute 4×A for E, and (2×A) for C in eq.4: 4×A = D ÷ (2×A) Multiply both sides of the above equation by (2×A): (2×A) × 4×A = (2×A) × D ÷ (2×A) which makes 8×A² = D
Hint #5
Substitute 8×A² for D, 4×A for E, (6×A) for B, and (2×A) for C in eq.5: 8×A² + 4×A = (6×A) × (2×A) which becomes 8×A² + 4×A = 12×A² Subtract 8×A² from each side of the above equation: 8×A² + 4×A – 8×A² = 12×A² – 8×A² which becomes eq.5a) 4×A = 4×A²
Solution
Since A ≠ 0 (2×A = C, and C ≠ 0 (from eq.4 and eq.6)), divide both sides of eq.5a by (4×A): 4×A ÷ (4×A) = 4×A² ÷ (4×A) which makes 1 = A and makes B = 6×A = 6 × 1 = 6 C = 2×A = 2 × 1 = 2 D = 8×A² = 8 × 1² = 8 × 1 = 8 E = 4×A = 4 × 1 = 4 F = A + D = 1 + 8 = 9 (from eq.1) and ABCDEF = 162849