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1354021778041 is a prime number
BaseRepresentation
bin10011101101000001111…
…111001011001001111001
311210102222002212010011021
4103231001333023021321
5134141014033344131
62514010133502441
7166552631133502
oct23550177131171
94712862763137
101354021778041
11482267aa3649
1219a5031a1a21
139a8b714238a
144976c09bba9
152534b8e2e11
hex13b41fcb279

1354021778041 has 2 divisors, whose sum is σ = 1354021778042. Its totient is φ = 1354021778040.

The previous prime is 1354021777963. The next prime is 1354021778051. The reversal of 1354021778041 is 1408771204531.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 847688490000 + 506333288041 = 920700^2 + 711571^2 .

It is a cyclic number.

It is not a de Polignac number, because 1354021778041 - 217 = 1354021646969 is a prime.

It is a super-2 number, since 2×13540217780412 (a number of 25 digits) contains 22 as substring.

It is not a weakly prime, because it can be changed into another prime (1354021778051) 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, 677010889020 + 677010889021.

It is an arithmetic number, because the mean of its divisors is an integer number (677010889021).

Almost surely, 21354021778041 is an apocalyptic number.

It is an amenable number.

1354021778041 is a deficient number, since it is larger than the sum of its proper divisors (1).

1354021778041 is an equidigital number, since it uses as much as digits as its factorization.

1354021778041 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 188160, while the sum is 43.

The spelling of 1354021778041 in words is "one trillion, three hundred fifty-four billion, twenty-one million, seven hundred seventy-eight thousand, forty-one".