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325020333437 is a prime number
BaseRepresentation
bin1001011101011001011…
…10001011010101111101
31011001221020002122210022
410232230232023111331
520311120201132222
6405151241421525
732324210152022
oct4565456132575
91131836078708
10325020333437
11115927561118
1252ba879a8a5
1324859634237
1411a3409d549
1586c404b042
hex4bacb8b57d

325020333437 has 2 divisors, whose sum is σ = 325020333438. Its totient is φ = 325020333436.

The previous prime is 325020333427. The next prime is 325020333457. The reversal of 325020333437 is 734333020523.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 244531239001 + 80489094436 = 494501^2 + 283706^2 .

It is an emirp because it is prime and its reverse (734333020523) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 325020333437 - 224 = 325003556221 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 325020333397 and 325020333406.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (325020333427) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (23) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 162510166718 + 162510166719.

It is an arithmetic number, because the mean of its divisors is an integer number (162510166719).

Almost surely, 2325020333437 is an apocalyptic number.

It is an amenable number.

325020333437 is a deficient number, since it is larger than the sum of its proper divisors (1).

325020333437 is an equidigital number, since it uses as much as digits as its factorization.

325020333437 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 136080, while the sum is 35.

The spelling of 325020333437 in words is "three hundred twenty-five billion, twenty million, three hundred thirty-three thousand, four hundred thirty-seven".