How many digits are in the smallest positive number divisible by th first ten numbers in the Fibonacci sequence?

first 10 Fibo numbers:

1 1 2 3 5 8 13 21 34 55

55 = 5x11
34 = 2x17
21 = 3x7
13 = 13
8 = 2x2x2
5 = 5

LCM = 17x11x13x7x5x3x2x2x2 = 2042040 , 7 digits

To find the smallest positive number divisible by the first ten numbers in the Fibonacci sequence, we need to calculate the least common multiple (LCM) of these ten numbers.

First, let's find the first ten numbers in the Fibonacci sequence:
1, 1, 2, 3, 5, 8, 13, 21, 34, 55

To find the LCM, we'll use the prime factorization method.

Let's prime-factorize each number in the Fibonacci sequence:

1 = 1^1
2 = 2^1
3 = 3^1
5 = 5^1
8 = 2^3
13 = 13^1
21 = 3^1 * 7^1
34 = 2^1 * 17^1
55 = 5^1 * 11^1

Now, we take the maximum exponent for each prime factor:

2^3 * 3^1 * 5^1 * 7^1 * 11^1 * 13^1 * 17^1

Calculating this product, we get:

2^3 * 3^1 * 5^1 * 7^1 * 11^1 * 13^1 * 17^1 = 510,510

Therefore, the smallest positive number divisible by the first ten numbers in the Fibonacci sequence is 510,510.

To determine the number of digits in this number, we can simply count the number of digits.

510,510 has 6 digits.

So, the smallest positive number divisible by the first ten numbers in the Fibonacci sequence has 6 digits.