How many natural numbers below 50 are there in the condition?

Question

that satisfying (n-1)! is not a multiple of n ?
1) 15
2) 16
3) 17
4) 18

please explain in detail

Answer 1

This will only be true for 4 and the prime numbers 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, and 47. Your answer is 16.

There are four categories of numbers to consider:

1) prime numbers: (x - 1)! does not include the prime number, so this must be true.

2) 1: 0! = 1, so the condition is false

3) other perfect squares: x = y * y. In this case, for the answer to be false, (x - 1)! must include both y and another multiple of y, such as 2y. This happens for every perfect square except 4.

4) every other number: you can write any other number as a product, x = y * z, where y != z and y < x, z < x. Both y and z are in (x - 1)!, so the condition is false.

Answer 2

I get 16 - answer number 2.
Here's why: in general in n is prime, then n will not divide (n-1)!, and the primes below 50 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, and 47. That's 15 in all. But the reason that n will generally divide (n-1)! if n is not prime is that the factors of n will appear in the factorial, and that can break down for small n.

For example 3! = 6 is not divisible by 4, so we've got to add 4 to the list. Now 5! = 120 is divisible by 6, and 7! contains factors of 2 and 4 so is divisible by 8. So 4 is the only problem, and we add that to the 15 primes for a total of 16 integers less than 50 which meet the criterion.

Answer 3

This will only be true for 4 and the prime numbers 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, and 47. Your answer is 16.

There are four categories of numbers to consider:

1) prime numbers: (x - 1)! does not include the prime number, so this must be true.

2) 1: 0! = 1, so the condition is false

3) other perfect squares: x = y * y. In this case, for the answer to be false, (x - 1)! must include both y and another multiple of y, such as 2y. This happens for every perfect square except 4.

4) every other number: you can write any other number as a product, x = y * z, where y != z and y < x, z < x. Both y and z are in (x - 1)!, so the condition is false.

Answer 4

1 & 4