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13011209752153 is a prime number
BaseRepresentation
bin1011110101010110100010…
…0010010000111001011001
31201001212012120121221101011
42331111220202100321121
53201133424144032103
643401133420253521
72512012623046504
oct275255042207131
951055176557334
1013011209752153
114167028781478
1215617ab9a12a1
13734c4b1a2b62
1432da62ad613b
151786b8859d6d
hexbd568890e59

13011209752153 has 2 divisors, whose sum is σ = 13011209752154. Its totient is φ = 13011209752152.

The previous prime is 13011209752069. The next prime is 13011209752271. The reversal of 13011209752153 is 35125790211031.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 11487239254089 + 1523970498064 = 3389283^2 + 1234492^2 .

It is a cyclic number.

It is not a de Polignac number, because 13011209752153 - 221 = 13011207655001 is a prime.

It is not a weakly prime, because it can be changed into another prime (13011209759153) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6505604876076 + 6505604876077.

It is an arithmetic number, because the mean of its divisors is an integer number (6505604876077).

Almost surely, 213011209752153 is an apocalyptic number.

It is an amenable number.

13011209752153 is a deficient number, since it is larger than the sum of its proper divisors (1).

13011209752153 is an equidigital number, since it uses as much as digits as its factorization.

13011209752153 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 56700, while the sum is 40.

Adding to 13011209752153 its reverse (35125790211031), we get a palindrome (48136999963184).

The spelling of 13011209752153 in words is "thirteen trillion, eleven billion, two hundred nine million, seven hundred fifty-two thousand, one hundred fifty-three".