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100131221093737 is a prime number
BaseRepresentation
bin10110110001000110011101…
…110111010011110101101001
3111010112110011221221202121111
4112301012131313103311221
5101111022220044444422
6552543351542110321
730043146001060531
oct2661063567236551
9433473157852544
10100131221093737
11299a545998a089
12b29212741b3a1
1343b4444cb9142
141aa253c53c4c1
15b89999938a77
hex5b119ddd3d69

100131221093737 has 2 divisors, whose sum is σ = 100131221093738. Its totient is φ = 100131221093736.

The previous prime is 100131221093707. The next prime is 100131221093741. The reversal of 100131221093737 is 737390122131001.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 68858397225216 + 31272823868521 = 8298096^2 + 5592211^2 .

It is a cyclic number.

It is not a de Polignac number, because 100131221093737 - 211 = 100131221091689 is a prime.

It is a super-2 number, since 2×1001312210937372 (a number of 29 digits) contains 22 as substring.

It is a self number, because there is not a number n which added to its sum of digits gives 100131221093737.

It is not a weakly prime, because it can be changed into another prime (100131221093707) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 50065610546868 + 50065610546869.

It is an arithmetic number, because the mean of its divisors is an integer number (50065610546869).

Almost surely, 2100131221093737 is an apocalyptic number.

It is an amenable number.

100131221093737 is a deficient number, since it is larger than the sum of its proper divisors (1).

100131221093737 is an equidigital number, since it uses as much as digits as its factorization.

100131221093737 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 47628, while the sum is 40.

The spelling of 100131221093737 in words is "one hundred trillion, one hundred thirty-one billion, two hundred twenty-one million, ninety-three thousand, seven hundred thirty-seven".