A is the set of positive integers such that when divided by 2, 3, 4, 5, 6 leaves remainders 1, 2, 3, 4.5 respectively. How many integers between 0 and 100 belong to set A?

Options

A. 0
B. 1
C. 2
D. none of these

Answer

To solve the problem, we need to find the set of positive integers \( A \) such that when divided by 2, 3, 4, 5, and 6, it leaves remainders of 1, 2, 3, 4, and 5 respectively. We will follow these steps:

### Step 1: Understand the problem
We need to find integers \( x \) such that:
- \( x \equiv 1 \mod 2 \)
- \( x \equiv 2 \mod 3 \)
- \( x \equiv 3 \mod 4 \)
- \( x \equiv 4 \mod 5 \)
- \( x \equiv 5 \mod 6 \)

### Step 2: Rewrite the congruences
We can rewrite the congruences in a more manageable form:
- \( x \equiv -1 \mod 2 \) (or \( x \equiv 1 \mod 2 \))
- \( x \equiv -1 \mod 3 \) (or \( x \equiv 2 \mod 3 \))
- \( x \equiv -1 \mod 4 \) (or \( x \equiv 3 \mod 4 \))
- \( x \equiv -1 \mod 5 \) (or \( x \equiv 4 \mod 5 \))
- \( x \equiv -1 \mod 6 \) (or \( x \equiv 5 \mod 6 \))

This means that \( x + 1 \) must be divisible by 2, 3, 4, 5, and 6.

### Step 3: Find the LCM
We need to find the least common multiple (LCM) of the numbers 2, 3, 4, 5, and 6:
- The prime factorization is:
- \( 2 = 2^1 \)
- \( 3 = 3^1 \)
- \( 4 = 2^2 \)
- \( 5 = 5^1 \)
- \( 6 = 2^1 \times 3^1 \)

The LCM takes the highest power of each prime:
- \( LCM = 2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 = 60 \)

### Step 4: Set up the equation
From the above, we have:
\[ x + 1 = 60k \]
for some integer \( k \). Thus,
\[ x = 60k - 1 \]

### Step 5: Find values of \( x \) between 0 and 100
We need to find \( k \) such that \( 0 < x < 100 \):
\[ 0 < 60k - 1 < 100 \]

Adding 1 to all parts:
\[ 1 < 60k < 101 \]

Dividing by 60:
\[ \frac{1}{60} < k < \frac{101}{60} \]

Calculating the bounds:
- \( \frac{1}{60} \approx 0.0167 \)
- \( \frac{101}{60} \approx 1.6833 \)

Thus, \( k \) can take the integer values \( 1 \).

### Step 6: Calculate \( x \) for valid \( k \)
For \( k = 1 \):
\[ x = 60 \times 1 - 1 = 59 \]

### Step 7: Check if there are other values
For \( k = 2 \):
\[ x = 60 \times 2 - 1 = 119 \] (which is greater than 100)

### Conclusion
The only integer in the set \( A \) between 0 and 100 is \( 59 \).

### Final Answer
There is **1 integer** in set \( A \) between 0 and 100.

---