Question

M is a 4 digit number. If the leftmost digit is removed, then the resulting three digit number is \(\frac{1}{9}^{th}\) of M. how many such M’s are possible ?

• A. 10
• B. 9
• C. 8
• D. 7

Answer 1

The Correct Option is D

To solve this problem, let's break down the information given:

1. M is a 4-digit number, so we can represent it as \(\overline{abcd}\), where 'a' is the leftmost digit, and is non-zero (since M is a 4-digit number).
2. When the leftmost digit is removed, the resulting 3-digit number becomes \(\overline{bcd}\).
3. It is given that \(\overline{bcd} = \frac{1}{9}\)th of M, i.e., \(\overline{bcd} = \frac{M}{9}\).

Let's denote M = 1000a + 100b + 10c + d and the resulting three-digit number as \(\overline{bcd} = 100b + 10c + d\).

The equation becomes:

100b + 10c + d = \frac{1000a + 100b + 10c + d}{9}

Multiply both sides by 9 to eliminate the fraction:

9(100b + 10c + d) = 1000a + 100b + 10c + d

This simplifies to:

900b + 90c + 9d = 1000a + 100b + 10c + d

Reorganize the equation:

1000a = 800b + 80c + 8d

Divide throughout by 8 to simplify:

125a = 100b + 10c + d

Based on this equation, for 'a' being a non-zero digit (1 to 9), we will check for valid values of \((b, c, d)\) that make \(\overline{bcd}\) satisfy the condition.

• If \(a = 1\): \(125 \times 1 = 125\). Valid \(\overline{bcd} = 125\).
• If \(a = 2\): \(125 \times 2 = 250\). Valid \(\overline{bcd} = 250\).
• If \(a = 3\): \(125 \times 3 = 375\). Valid \(\overline{bcd} = 375\).
• If \(a = 4\): \(125 \times 4 = 500\). Valid \(\overline{bcd} = 500\).
• If \(a = 5\): \(125 \times 5 = 625\). Valid \(\overline{bcd} = 625\).
• If \(a = 6\): \(125 \times 6 = 750\). Valid \(\overline{bcd} = 750\).
• If \(a = 7\): \(125 \times 7 = 875\). Valid \(\overline{bcd} = 875\).

Therefore, for each value of 'a' (from 1 to 7), the possible numbers M are 1125, 2250, 3375, 4500, 5625, 6750, and 7875, respectively. There are 7 such numbers M possible.

Conclusion: The correct option is 7.