The largest number of five digits which, when divided by 16, 24, 30 and 36 leaves the same  remainder 10 in each case is

Explanation

First, we need to calculate the least common multiple (LCM) of the divisors (16, 24, 30, and 36). The LCM will help us find a common base for our calculations.

Prime factorization:

• 16 = 2⁴
• 24 = 2³ × 3
• 30 = 2 × 3 × 5
• 36 = 2² × 3²

Now, take the highest power of each prime:

• For 2: 2⁴
• For 3: 3²
• For 5: 5¹

Thus, the LCM is:

LCM = 2⁴ × 3² × 5 = 16 × 9 × 5 = 720

The largest five-digit number is 99999. To find a number that leaves a remainder of 10 when divided by our calculated LCM (720), we can express it as:

N = k × LCM + R

where R = 10.

We want:

N = k × 720 + 10

To find the largest N < 99999:

k × 720 + 10 < 99999

Subtracting 10 from both sides gives:

k × 720 < 99989

Now divide by 720:

k < &frac{99989}{720} ≈ 138.88

Thus, the largest integer value for k is 138.

Now substitute back to find N:

N = 138 × 720 + 10 = 99360 + 10 = 99370

Final Answer

The largest five-digit number that leaves a remainder of 10 when divided by 16, 24, 30, and 36, is: b. 99370