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99294104773 is a prime number
BaseRepresentation
bin101110001111001100…
…0111100110011000101
3100111021222101011001001
41130132120330303011
53111323242323043
6113340505133301
710113412651633
oct1343630746305
9314258334031
1099294104773
1139123a2a616
12172b148a231
1394954b7c66
144b3d3d0353
1528b22a074d
hex171e63ccc5

99294104773 has 2 divisors, whose sum is σ = 99294104774. Its totient is φ = 99294104772.

The previous prime is 99294104737. The next prime is 99294104783. The reversal of 99294104773 is 37740149299.

Together with previous prime (99294104737) it forms an Ormiston pair, because they use the same digits, order apart.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 96129142209 + 3164962564 = 310047^2 + 56258^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-99294104773 is a prime.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (99294104783) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 49647052386 + 49647052387.

It is an arithmetic number, because the mean of its divisors is an integer number (49647052387).

Almost surely, 299294104773 is an apocalyptic number.

It is an amenable number.

99294104773 is a deficient number, since it is larger than the sum of its proper divisors (1).

99294104773 is an equidigital number, since it uses as much as digits as its factorization.

99294104773 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 3429216, while the sum is 55.

The spelling of 99294104773 in words is "ninety-nine billion, two hundred ninety-four million, one hundred four thousand, seven hundred seventy-three".