Q65 / 180 · COMEDK 2025 (10 May Shift 1) · Mathematics
The digits of a three-digit number taken in an order are in geometric progression. If one is added to the middle digit, they form an arithmetic progression. If 594 is subtracted from the number, then a new number with the same digits in reverse order is formed. The original number is divisible by
- A$4$
- B$19$
- C$11$
- Correct answerD$421$
Solution
Let the digits be $a, ar, ar^2$ in GP. The number is $100a + 10ar + ar^2$.
AP condition ($a,\ b+1,\ c$ in AP): $2(ar+1) = a + ar^2 \Rightarrow 2ar + 2 = a(1+r^2)$ ... (1)
Reverse condition: $(100a + 10ar + ar^2) - 594 = 100ar^2 + 10ar + a$
$99a - 99ar^2 = 594 \Rightarrow a(1-r^2) = 6$ ... (2)
From (1): $a - ar^2 = 2 - 2ar$. Substituting into (2): $2 - 2ar = 6 \Rightarrow ar = -2$
Since digits must be positive and $a > c$ (as 594 is subtracted), try $r = \dfrac{1}{2}$, $a = 8$:
$b = 8 \times \dfrac{1}{2} = 4$, $c = 8 \times \dfrac{1}{4} = 2$
Verification: $8, 4, 2$ are in GP. $8, 5, 2$ are in AP. $842 - 594 = 248$ (reverse of 842).
The number is $842 = 2 \times 421$, so it is divisible by $421$.
AP condition ($a,\ b+1,\ c$ in AP): $2(ar+1) = a + ar^2 \Rightarrow 2ar + 2 = a(1+r^2)$ ... (1)
Reverse condition: $(100a + 10ar + ar^2) - 594 = 100ar^2 + 10ar + a$
$99a - 99ar^2 = 594 \Rightarrow a(1-r^2) = 6$ ... (2)
From (1): $a - ar^2 = 2 - 2ar$. Substituting into (2): $2 - 2ar = 6 \Rightarrow ar = -2$
Since digits must be positive and $a > c$ (as 594 is subtracted), try $r = \dfrac{1}{2}$, $a = 8$:
$b = 8 \times \dfrac{1}{2} = 4$, $c = 8 \times \dfrac{1}{4} = 2$
Verification: $8, 4, 2$ are in GP. $8, 5, 2$ are in AP. $842 - 594 = 248$ (reverse of 842).
The number is $842 = 2 \times 421$, so it is divisible by $421$.
Hence, the correct answer is option: 'D'.