To solve the problem step by step, we need to find the largest six-digit number that, when divided by 12, 15, 20, 24, and 30, leaves specific remainders.
### Step 1: Identify the largest six-digit number
The largest six-digit number is 999999.
### Step 2: Determine the remainders
The problem states that:
- When divided by 12, the remainder is 8.
- When divided by 15, the remainder is 11.
- When divided by 20, the remainder is 16.
- When divided by 24, the remainder is 20.
- When divided by 30, the remainder is 26.
### Step 3: Convert the remainders
We can convert the problem into a form that is easier to handle. For each divisor, we can express the conditions as:
- \(999999 - 8\) is divisible by 12
- \(999999 - 11\) is divisible by 15
- \(999999 - 16\) is divisible by 20
- \(999999 - 20\) is divisible by 24
- \(999999 - 26\) is divisible by 30
This means that:
- \(999999 - 8 \equiv 0 \mod 12\)
- \(999999 - 11 \equiv 0 \mod 15\)
- \(999999 - 16 \equiv 0 \mod 20\)
- \(999999 - 20 \equiv 0 \mod 24\)
- \(999999 - 26 \equiv 0 \mod 30\)
### Step 4: Find a common value
From the above, we can see that:
- \(999999 - 8 \equiv 0 \mod 12\) implies \(999991 \equiv 0 \mod 12\)
- \(999999 - 11 \equiv 0 \mod 15\) implies \(999988 \equiv 0 \mod 15\)
- \(999999 - 16 \equiv 0 \mod 20\) implies \(999983 \equiv 0 \mod 20\)
- \(999999 - 20 \equiv 0 \mod 24\) implies \(999979 \equiv 0 \mod 24\)
- \(999999 - 26 \equiv 0 \mod 30\) implies \(999973 \equiv 0 \mod 30\)
### Step 5: Calculate the LCM
We need to find the least common multiple (LCM) of the divisors: 12, 15, 20, 24, and 30.
- The prime factorization of each number is:
- 12 = \(2^2 \times 3\)
- 15 = \(3 \times 5\)
- 20 = \(2^2 \times 5\)
- 24 = \(2^3 \times 3\)
- 30 = \(2 \times 3 \times 5\)
The LCM is calculated by taking the highest power of each prime:
- LCM = \(2^3 \times 3^1 \times 5^1 = 120\)
### Step 6: Find the largest number that meets the conditions
Now we need to find the largest six-digit number that is less than or equal to 999999 and is of the form \(999999 - k\), where \(k\) is a multiple of 120.
To find the largest multiple of 120 less than 999999, we perform:
\[
999999 \div 120 = 8333.325
\]
Taking the integer part, we get 8333. Thus, the largest multiple of 120 is:
\[
120 \times 8333 = 999960
\]
### Step 7: Adjust for remainders
Now, we need to adjust for the remainders:
Since we need a number that leaves a remainder of 4 when divided by each of the divisors, we subtract 4 from 999960:
\[
999960 - 4 = 999956
\]
### Final Answer
Thus, the largest six-digit number that meets all the conditions is **999956**.
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