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1157010091837 is a prime number
BaseRepresentation
bin10000110101100011001…
…011001110111100111101
311002121110000121111112121
4100311203023032330331
5122424024040414322
62243304553402541
7146406534340555
oct20654313167475
94077400544477
101157010091837
1140675a040876
121682a0409451
138514ac75522
143dddcc20165
152016a9a4ec7
hex10d632cef3d

1157010091837 has 2 divisors, whose sum is σ = 1157010091838. Its totient is φ = 1157010091836.

The previous prime is 1157010091747. The next prime is 1157010091871. The reversal of 1157010091837 is 7381900107511.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 688358946276 + 468651145561 = 829674^2 + 684581^2 .

It is a cyclic number.

It is not a de Polignac number, because 1157010091837 - 211 = 1157010089789 is a prime.

It is a super-2 number, since 2×11570100918372 (a number of 25 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 1157010091793 and 1157010091802.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (1157010091337) 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, 578505045918 + 578505045919.

It is an arithmetic number, because the mean of its divisors is an integer number (578505045919).

Almost surely, 21157010091837 is an apocalyptic number.

It is an amenable number.

1157010091837 is a deficient number, since it is larger than the sum of its proper divisors (1).

1157010091837 is an equidigital number, since it uses as much as digits as its factorization.

1157010091837 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 52920, while the sum is 43.

The spelling of 1157010091837 in words is "one trillion, one hundred fifty-seven billion, ten million, ninety-one thousand, eight hundred thirty-seven".