Question

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. 19
• B. 11
• C. 421
• D. 4

Hint:

For digit-based number problems, assume the digits as variables and convert each given condition into an algebraic equation.

Answer 1

The Correct Option is C

Step 1: Assume the digits of the number.
Let the three-digit number be:
\[ 100a+10b+c \]
where \(a\), \(b\), and \(c\) are its digits.

Step 2: Use the condition of geometric progression.
Since the digits are in geometric progression, we have:
\[ b^2=ac. \]

Step 3: Use the arithmetic progression condition.
If one is added to the middle digit, the digits become:
\[ a,\ b+1,\ c. \]
These are in arithmetic progression, so:
\[ 2(b+1)=a+c. \]

Step 4: Use the reverse number condition.
When 594 is subtracted from the original number, the digits are reversed:
\[ 100a+10b+c-594=100c+10b+a. \]
Simplifying:
\[ 99a-99c=594. \]
\[ 99(a-c)=594. \]
\[ a-c=6. \]
So,
\[ a=c+6. \]

Step 5: Substitute in the AP condition.
Using:
\[ 2(b+1)=a+c, \]
and \(a=c+6\), we get:
\[ 2b+2=c+6+c. \]
\[ 2b+2=2c+6. \]
\[ 2b=2c+4. \]
\[ b=c+2. \]

Step 6: Substitute in the GP condition.
Using \(b^2=ac\), \(a=c+6\), and \(b=c+2\), we get:
\[ (c+2)^2=c(c+6). \]
Expanding both sides:
\[ c^2+4c+4=c^2+6c. \]
Cancel \(c^2\) from both sides:
\[ 4c+4=6c. \]
\[ 4=2c. \]
\[ c=2. \]
Therefore:
\[ b=c+2=4, \] and
\[ a=c+6=8. \]

Step 7: Find the original number and divisibility.
The original number is:
\[ 100a+10b+c=100(8)+10(4)+2. \]
\[ =800+40+2=842. \]
Now,
\[ 842=2 \times 421. \]
So, the original number is divisible by \(421\).
Final Answer:
\[ \boxed{421}. \]