Find the least number that must be added to 1000 so that the sum is exactly divisible by 16, 64 and 256.

Answer

To find the least number that must be added to 1000 so that the sum is exactly divisible by 16, 64, and 256, we will follow these steps:

### Step 1: Find the LCM of 16, 64, and 256
To determine the least common multiple (LCM), we can use the prime factorization method.

- The prime factorization of 16 is \(2^4\).
- The prime factorization of 64 is \(2^6\).
- The prime factorization of 256 is \(2^8\).

The LCM is found by taking the highest power of each prime factor present in the numbers.

Thus, the LCM of 16, 64, and 256 is \(2^8 = 256\).

### Step 2: Divide 1000 by the LCM
Next, we will divide 1000 by the LCM (256) to find out how many times it fits into 1000.

\[ 1000 \div 256 \approx 3.90625 \]

This means 256 fits into 1000 a total of 3 times.

### Step 3: Calculate the product of the LCM and the quotient
Now, we will multiply the LCM by the whole number part of the division result:

\[ 256 \times 3 = 768 \]

### Step 4: Find the remainder
Now, we will find the remainder when 1000 is subtracted by this product:

\[ 1000 - 768 = 232 \]

### Step 5: Calculate the least number to be added
To make 1000 divisible by 256, we need to add a number that will make the total equal to the next multiple of 256. We can find this by subtracting the remainder from the LCM:

\[ 256 - 232 = 24 \]

Thus, the least number that must be added to 1000 so that the sum is exactly divisible by 16, 64, and 256 is **24**.

### Summary of the Solution
The least number to be added to 1000 is **24**.

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