What least number must be subtracted from 2963 so that the resulting number when divided by 9, 10 and 15, the remainder in each case is 5?

Options

A. 41
B. 78
C. 82
D. 39

Answer

To solve the problem, we need to find the least number that must be subtracted from 2963 so that the resulting number gives a remainder of 5 when divided by 9, 10, and 15.

### Step-by-Step Solution:

1. **Understand the Requirement**:
We need a number \( x \) such that when \( 2963 - x \) is divided by 9, 10, and 15, the remainder is 5. This means:
\[ 2963 - x \equiv 5 \mod 9 \]
\[ 2963 - x \equiv 5 \mod 10 \]
\[ 2963 - x \equiv 5 \mod 15 \]

2. **Rearranging the Congruences**:
We can rearrange the above congruences:
\[ 2963 - 5 \equiv x \mod 9 \]
\[ 2963 - 5 \equiv x \mod 10 \]
\[ 2963 - 5 \equiv x \mod 15 \]
Which simplifies to:
\[ x \equiv 2958 \mod 9 \]
\[ x \equiv 2958 \mod 10 \]
\[ x \equiv 2958 \mod 15 \]

3. **Finding the Least Common Multiple (LCM)**:
To find a common solution for \( x \), we first find the LCM of the divisors (9, 10, and 15):
- The prime factorization is:
- \( 9 = 3^2 \)
- \( 10 = 2 \times 5 \)
- \( 15 = 3 \times 5 \)
- The LCM is:
\[ LCM = 2^1 \times 3^2 \times 5^1 = 90 \]

4. **Finding the Remainder**:
Now we need to find \( 2958 \mod 90 \):
\[ 2958 \div 90 = 32 \quad \text{(integer part)} \]
\[ 32 \times 90 = 2880 \]
\[ 2958 - 2880 = 78 \]
Thus, \( 2958 \equiv 78 \mod 90 \).

5. **Calculating the Number to be Subtracted**:
Since we need \( x \equiv 78 \), we can now find the least number to subtract from 2963:
\[ x = 2963 - 78 = 2885 \]

### Final Answer:
The least number that must be subtracted from 2963 is **78**.