All numbers ≤4 divide 10 (the base of the numeral system of this wiki), but 5 does not.

All numbers ≤4 satisfy the property that the algebraic equations with order n have algebraic solution, but 5 does not.

All numbers ≤4 satisfy the property that the Fermat number 2^(2^n)+1 is prime, but 5 does not.

All numbers ≤4 satisfy the property that the double Mersenne number 2^(2^(nth prime)−1)−1 is prime, but 5 does not.

All numbers ≤4 satisfy the property that the completed graph K_n is planar graph, but 5 does not.

All numbers ≤4 satisfy the property that 1/n has terminated dozenal, but 5 does not.

All numbers ≤4 satisfy the property that the group A_n is a solvable group, but 5 does not.

All numbers ≤4 satisfy the property that the Lucas number L(2^n) is prime, but 5 does not.

For the number 4! = 20:

If and only if *n* is a divisor of 20, then *m*^{2} = 1 mod *n* for every integer *m* coprime to *n*.

If and only if *n* is a divisor of 20, then the Dirichlet characters mod *n* are all real.

If and only if *n* is a divisor of 20, then *n* is divisible by all numbers less than or equal to the square root of *n*.

If and only if *n* is a divisor of 20, then *k*−1 is prime for all divisors *k*>2 of *n*.

If and only if *n*+1 is a divisor of 20, then $ \tbinom{n}{k}=\tfrac{n!}{k!(n-k)!} $ is squarefree for all 0 ≤ *k* ≤ *n*, i.e. all numbers in the *n*th row of the Pascal's triangle are squarefree (the topmost row (i.e. the row which contains only one 1) of the Pascal's triangle is the 0th row, not the 1st row). (Note that all such *n* are primes or 1 or 0, and 20 is the largest number *m* such that if *n*+1 is a divisor of *m*, then *n* is prime or 1 or 0, besides, if and only if *m* is a divisor of 20, then *m* satisfies this condition)