Which is the largest five digit number, which when divided by 6, 8, 15, 20 and 30 leaves the remainders 2, 4, 11, 16 and 26 respectively.

Options

A. 99986
B. 99956
C. 999546
D. 99812

Answer

To solve the problem of finding the largest five-digit number that leaves specific remainders when divided by 6, 8, 15, 20, and 30, we can follow these steps:

### Step 1: Understand the Problem
We need to find a number \( N \) such that:
- \( N \mod 6 = 2 \)
- \( N \mod 8 = 4 \)
- \( N \mod 15 = 11 \)
- \( N \mod 20 = 16 \)
- \( N \mod 30 = 26 \)

### Step 2: Convert the Remainders
To simplify the problem, we can rewrite the conditions in terms of \( N \):
- \( N = 6k + 2 \)
- \( N = 8m + 4 \)
- \( N = 15n + 11 \)
- \( N = 20p + 16 \)
- \( N = 30q + 26 \)

This implies that:
- \( N - 2 \) is divisible by 6
- \( N - 4 \) is divisible by 8
- \( N - 11 \) is divisible by 15
- \( N - 16 \) is divisible by 20
- \( N - 26 \) is divisible by 30

### Step 3: Find the LCM
To find a common number that satisfies all these conditions, we need to find the least common multiple (LCM) of the divisors:
- The divisors are 6, 8, 15, 20, and 30.

Calculating the LCM:
- Prime factorization:
- \( 6 = 2 \times 3 \)
- \( 8 = 2^3 \)
- \( 15 = 3 \times 5 \)
- \( 20 = 2^2 \times 5 \)
- \( 30 = 2 \times 3 \times 5 \)

Taking the highest powers of each prime:
- \( 2^3 \) from 8
- \( 3^1 \) from 6 or 15
- \( 5^1 \) from 15 or 20

Thus, \( \text{LCM} = 2^3 \times 3^1 \times 5^1 = 8 \times 3 \times 5 = 120 \).

### Step 4: Find the Largest Five-Digit Number
The largest five-digit number is 99999. We need to find the largest multiple of 120 that is less than or equal to 99999.

To find this, we divide:
\[ 99999 \div 120 \approx 833.325 \]
Taking the integer part, we have:
\[ 833 \]

Now, we calculate:
\[ 833 \times 120 = 99960 \]

### Step 5: Adjust for the Remainders
Now we need to adjust \( 99960 \) to ensure it meets the original remainder conditions:
- Since \( N \equiv 2 \mod 6 \), we check \( 99960 \mod 6 \):
- \( 99960 \mod 6 = 0 \) (not correct)

We need to subtract \( 4 \) (since \( 2 \) is the desired remainder):
\[ 99960 - 4 = 99956 \]

### Step 6: Verify the Result
Now we check if \( 99956 \) satisfies all the original conditions:
- \( 99956 \mod 6 = 2 \) (correct)
- \( 99956 \mod 8 = 4 \) (correct)
- \( 99956 \mod 15 = 11 \) (correct)
- \( 99956 \mod 20 = 16 \) (correct)
- \( 99956 \mod 30 = 26 \) (correct)

### Conclusion
The largest five-digit number that satisfies all the conditions is:
\[ \boxed{99956} \]