Question

Is there a number divisible by 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10

Answer 1

Yes, there is a number that is divisible by all of the integers from 1 to 10, and that number is the least common multiple (LCM) of those integers. To find the LCM of 1 through 10, we can factor each number into its prime factors:

- \(1\) has no prime factors.
- \(2 = 2\)
- \(3 = 3\)
- \(4 = 2^2\)
- \(5 = 5\)
- \(6 = 2 \times 3\)
- \(7 = 7\)
- \(8 = 2^3\)
- \(9 = 3^2\)
- \(10 = 2 \times 5\)

Next, we take the highest power of each prime that appears in these factorizations:

- The highest power of \(2\) is \(2^3\) (from \(8\)).
- The highest power of \(3\) is \(3^2\) (from \(9\)).
- The highest power of \(5\) is \(5^1\) (from \(5\) and \(10\)).
- The highest power of \(7\) is \(7^1\) (from \(7\)).

Now, we can calculate the LCM:

\[
\text{LCM} = 2^3 \times 3^2 \times 5^1 \times 7^1
\]

Calculating this step-by-step:

\[
2^3 = 8
\]
\[
3^2 = 9
\]
\[
5^1 = 5
\]
\[
7^1 = 7
\]

Now, multiply them together:

\[
8 \times 9 = 72
\]
\[
72 \times 5 = 360
\]
\[
360 \times 7 = 2520
\]

Thus, the least common multiple of the numbers from \(1\) to \(10\) is \(2520\).

So, \(2520\) is a number that is divisible by all of the integers from \(1\) to \(10\).