A number when divided by 5 gives a number which is 8 more than the remainder obtained on dividing the same number by 34. Such a least possible number is :

Options

A. 175
B. 75
C. 680
D. does not exist

Answer

To solve the problem step by step, let's define the number as \( x \).

### Step 1: Understand the conditions given in the problem.
The problem states that when \( x \) is divided by 5, the result is 8 more than the remainder when \( x \) is divided by 34.

### Step 2: Set up the equations based on the conditions.
1. When \( x \) is divided by 5, we can express this as:
\[ \frac{x}{5} = q_1 \quad \text{(where \( q_1 \) is the quotient)} \]
This implies:
\[ x = 5q_1 + r_1 \quad \text{(where \( r_1 \) is the remainder when \( x \) is divided by 5)} \]
Since the remainder when dividing by 5 can be 0, 1, 2, 3, or 4, we have:
\[ r_1 \in \{0, 1, 2, 3, 4\} \]

2. When \( x \) is divided by 34, we can express this as:
\[ x = 34q_2 + r_2 \quad \text{(where \( r_2 \) is the remainder when \( x \) is divided by 34)} \]
The remainder \( r_2 \) can be any integer from 0 to 33.

### Step 3: Relate the two conditions.
According to the problem:
\[ q_1 = r_2 + 8 \]
This means:
\[ \frac{x}{5} = r_2 + 8 \]

### Step 4: Substitute \( x \) in terms of \( r_2 \).
From the equation \( x = 5(r_2 + 8) \), we can express \( x \) as:
\[ x = 5r_2 + 40 \]

### Step 5: Find the remainder when \( x \) is divided by 34.
Now we need to check the remainder when \( x = 5r_2 + 40 \) is divided by 34:
\[ x \mod 34 = (5r_2 + 40) \mod 34 \]

### Step 6: Simplify the expression.
Calculating \( 40 \mod 34 \):
\[ 40 \mod 34 = 6 \]
Thus:
\[ x \mod 34 = (5r_2 + 6) \mod 34 \]

### Step 7: Set up the condition for \( r_2 \).
We know that \( r_2 \) must be less than 8 (since \( q_1 \) must be non-negative):
\[ r_2 < 8 \]
So \( r_2 \) can take values from 0 to 7.

### Step 8: Check for the least possible \( x \).
We will check values of \( r_2 \) from 0 to 7 to find the least possible \( x \):
- For \( r_2 = 0 \): \( x = 5(0) + 40 = 40 \)
- For \( r_2 = 1 \): \( x = 5(1) + 40 = 45 \)
- For \( r_2 = 2 \): \( x = 5(2) + 40 = 50 \)
- For \( r_2 = 3 \): \( x = 5(3) + 40 = 55 \)
- For \( r_2 = 4 \): \( x = 5(4) + 40 = 60 \)
- For \( r_2 = 5 \): \( x = 5(5) + 40 = 65 \)
- For \( r_2 = 6 \): \( x = 5(6) + 40 = 70 \)
- For \( r_2 = 7 \): \( x = 5(7) + 40 = 75 \)

### Step 9: Identify the least possible number.
The least possible number \( x \) that satisfies all conditions is:
\[ \boxed{75} \]