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5184425437 is a prime number
BaseRepresentation
bin1001101010000010…
…00000110111011101
3111101022102002200221
410311001000313131
541104203103222
62214240121341
7242321613652
oct46501006735
914338362627
105184425437
112220522624
121008306251
13648121b79
1437278bc29
152052360c7
hex135040ddd

5184425437 has 2 divisors, whose sum is σ = 5184425438. Its totient is φ = 5184425436.

The previous prime is 5184425429. The next prime is 5184425479. The reversal of 5184425437 is 7345244815.

5184425437 is digitally balanced in base 3, because in such base it contains all the possibile digits an equal number of times.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 3477578841 + 1706846596 = 58971^2 + 41314^2 .

It is a cyclic number.

It is not a de Polignac number, because 5184425437 - 23 = 5184425429 is a prime.

It is a super-2 number, since 2×51844254372 = 53756534223625281938, which contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 5184425393 and 5184425402.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (5184425137) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2592212718 + 2592212719.

It is an arithmetic number, because the mean of its divisors is an integer number (2592212719).

Almost surely, 25184425437 is an apocalyptic number.

It is an amenable number.

5184425437 is a deficient number, since it is larger than the sum of its proper divisors (1).

5184425437 is an equidigital number, since it uses as much as digits as its factorization.

5184425437 is an odious number, because the sum of its binary digits is odd.

The product of its digits is 537600, while the sum is 43.

The square root of 5184425437 is about 72002.9543629982. The cubic root of 5184425437 is about 1730.7468276981.

The spelling of 5184425437 in words is "five billion, one hundred eighty-four million, four hundred twenty-five thousand, four hundred thirty-seven".