Finding the Highest Four-Digit Number Divisible by Multiple Numbers

The problem asks us to find the largest four-digit number that is perfectly divisible by several given numbers: 16, 36, 45, and 48. A number that is divisible by each of these numbers must be a multiple of their Least Common Multiple (LCM).

Step 1: Find the LCM of the Given Numbers

To find the LCM of 16, 36, 45, and 48, we first find the prime factorization of each number.

• Prime factorization of 16: \(\text{16} = \text{2} \times \text{2} \times \text{2} \times \text{2} = \text{2}^4\)
• Prime factorization of 36: \(\text{36} = \text{2} \times \text{2} \times \text{3} \times \text{3} = \text{2}^2 \times \text{3}^2\)
• Prime factorization of 45: \(\text{45} = \text{3} \times \text{3} \times \text{5} = \text{3}^2 \times \text{5}^1\)
• Prime factorization of 48: \(\text{48} = \text{2} \times \text{2} \times \text{2} \times \text{2} \times \text{3} = \text{2}^4 \times \text{3}^1\)

Now, we list all prime factors that appear in any of the factorizations and take the highest power of each factor:

• The prime factors are 2, 3, and 5.
• Highest power of 2 is \(2^4\) (from 16 and 48).
• Highest power of 3 is \(3^2\) (from 36 and 45).
• Highest power of 5 is \(5^1\) (from 45).

The LCM is the product of these highest powers:

\(\text{LCM(16, 36, 45, 48)} = \text{2}^4 \times \text{3}^2 \times \text{5}^1 = \text{16} \times \text{9} \times \text{5}\)

\(\text{LCM} = \text{144} \times \text{5} = \text{720}\)

So, any number divisible by 16, 36, 45, and 48 must be a multiple of 720.

Step 2: Find the Highest Four-Digit Multiple of the LCM

We are looking for the largest four-digit number that is a multiple of 720. The largest four-digit number is 9999.

We need to find the largest multiple of 720 that is less than or equal to 9999. We can do this by dividing 9999 by 720.

\(\text{9999} \div \text{720}\)

Using division:

The division gives a quotient of 13 and a remainder of 639.

This means \(9999 = 720 \times 13 + 639\).

The largest multiple of 720 less than or equal to 9999 is found by multiplying the quotient by the LCM:

\(\text{Largest multiple} = \text{720} \times \text{13}\)

\(\text{720} \times \text{13} = \text{9360}\)

The number 9360 is a four-digit number. The next multiple of 720 would be \(720 \times 14 = 9360 + 720 = 10080\), which is a five-digit number.

Conclusion: Highest Four-Digit Divisible Number

Therefore, the highest four-digit number that is divisible by 16, 36, 45, and 48 is 9360.

Revision Table: Divisibility and LCM Concepts

Additional Information: Finding Multiples

When you need to find the largest multiple of a number 'N' that is less than or equal to another number 'M', you can follow these steps:

1. Divide M by N. Calculate the quotient, let's call it Q, and the remainder, R. \(\text{M} = \text{N} \times \text{Q} + \text{R}\)
2. The largest multiple of N less than or equal to M is \(\text{N} \times \text{Q}\).

In our case, M = 9999 and N = 720. We found Q = 13 and R = 639. So, the largest multiple of 720 less than or equal to 9999 is \(720 \times 13 = 9360\). Another way to think about it is to subtract the remainder from the largest number: \(9999 - 639 = 9360\).