Question

Sujith says: - Multiplying first two digits by 3 gives all ones.
- Multiplying next two digits by 6 gives all twos.
- Multiplying last two digits by 9 gives all threes.
What is the sum of digits in the number?

• A. 30
• B. 33
• C. 60
• D. 45

Hint:

Use reverse arithmetic to reconstruct each pair, check consistency, and add digits.

Answer 1

The Correct Option is D

Let the 6-digit number be: \( \overline{AABBCC} \) Step 1: First two digits: \[ AB \times 3 = 111 \Rightarrow AB = \frac{111}{3} = 37 \Rightarrow A = 3, B = 7 \] Step 2: Next two digits: \[ BB \times 6 = 222 \Rightarrow BB = \frac{222}{6} = 37 \Rightarrow Same: A = 3, B = 7 \] Step 3: Last two digits: \[ CC \times 9 = 333 \Rightarrow CC = \frac{333}{9} = 37 \Rightarrow C = 7 So full number = 373737 Sum of digits: \( 3 + 7 + 3 + 7 + 3 + 7 = \boxed{30} \) Wait! But above result shows 3+7+3+7+3+7 = 30 → So Option (A)
But if we try:
111 / 3 = 37 → OK
222 / 6 = 37 → OK
333 / 9 = 37 → OK
So number is 373737 → digits are 3,7 repeated
Sum = \(3+7+3+7+3+7 = \boxed{34}\) → No option matches
Wait! Let's try:
Step-by-step reconstruction:
- \( x \times 3 = 111 \Rightarrow x = 37 \)
- \( x \times 6 = 222 \Rightarrow x = 37 \)
- \( x \times 9 = 333 \Rightarrow x = 37 \)
So all 3 parts are 37 → Number = 373737
Digits: 3,7,3,7,3,7
Sum = \(3 + 7 + 3 + 7 + 3 + 7 = \boxed{30}\)
Final Answer: \( \boxed{30} \)