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100102100000153 is a prime number
BaseRepresentation
bin10110110000101011010110…
…000111001001110110011001
3111010102200000110110221011212
4112300223112013021312121
5101110033100100001103
6552522134144032505
730041065232625206
oct2660532607116631
9433380013427155
10100102100000153
1129994075880499
12b28855a902735
1343b1783a43614
141aa0d78b1ccad
15b88d430717d8
hex5b0ad61c9d99

100102100000153 has 2 divisors, whose sum is σ = 100102100000154. Its totient is φ = 100102100000152.

The previous prime is 100102100000099. The next prime is 100102100000171. The reversal of 100102100000153 is 351000001201001.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 99692976021769 + 409123978384 = 9984637^2 + 639628^2 .

It is an emirp because it is prime and its reverse (351000001201001) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 100102100000153 - 28 = 100102099999897 is a prime.

It is a Sophie Germain prime.

It is a Curzon number.

It is not a weakly prime, because it can be changed into another prime (100102100002153) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 50051050000076 + 50051050000077.

It is an arithmetic number, because the mean of its divisors is an integer number (50051050000077).

Almost surely, 2100102100000153 is an apocalyptic number.

It is an amenable number.

100102100000153 is a deficient number, since it is larger than the sum of its proper divisors (1).

100102100000153 is an equidigital number, since it uses as much as digits as its factorization.

100102100000153 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 30, while the sum is 14.

Adding to 100102100000153 its reverse (351000001201001), we get a palindrome (451102101201154).

The spelling of 100102100000153 in words is "one hundred trillion, one hundred two billion, one hundred million, one hundred fifty-three", and thus it is an aban number.