Search a number
-
+
1300235455241 is a prime number
BaseRepresentation
bin10010111010111100000…
…100101000101100001001
311121022010122101211002122
4102322330010220230021
5132300340234031431
62433153034141025
7162640022004662
oct22727404505411
94538118354078
101300235455241
114614770211a8
1218bbb2281775
13957c4b16778
1446d0895a369
1523c4e934a7b
hex12ebc128b09

1300235455241 has 2 divisors, whose sum is σ = 1300235455242. Its totient is φ = 1300235455240.

The previous prime is 1300235455223. The next prime is 1300235455243. The reversal of 1300235455241 is 1425545320031.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 1155528252025 + 144707203216 = 1074955^2 + 380404^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-1300235455241 is a prime.

It is a super-2 number, since 2×13002354552412 (a number of 25 digits) contains 22 as substring.

Together with 1300235455243, it forms a pair of twin primes.

It is a Chen prime.

It is not a weakly prime, because it can be changed into another prime (1300235455243) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 650117727620 + 650117727621.

It is an arithmetic number, because the mean of its divisors is an integer number (650117727621).

Almost surely, 21300235455241 is an apocalyptic number.

It is an amenable number.

1300235455241 is a deficient number, since it is larger than the sum of its proper divisors (1).

1300235455241 is an equidigital number, since it uses as much as digits as its factorization.

1300235455241 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 72000, while the sum is 35.

The spelling of 1300235455241 in words is "one trillion, three hundred billion, two hundred thirty-five million, four hundred fifty-five thousand, two hundred forty-one".