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110000110000153 is a prime number
BaseRepresentation
bin11001000000101101100101…
…011110110101100000011001
3112102110220111210112100221211
4121000231211132311200121
5103404220211130001103
61025541204121303121
732112150161511652
oct3100554536654031
9472426453470854
10110000110000153
113205a86a441a0a
1210406912334aa1
13494bc730068c9
141d240687b6529
15cab54e1e036d
hex640b657b5819

110000110000153 has 2 divisors, whose sum is σ = 110000110000154. Its totient is φ = 110000110000152.

The previous prime is 110000110000097. The next prime is 110000110000201. The reversal of 110000110000153 is 351000011000011.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 100323300887104 + 9676809113049 = 10016152^2 + 3110757^2 .

It is a cyclic number.

It is not a de Polignac number, because 110000110000153 - 233 = 109991520065561 is a prime.

It is not a weakly prime, because it can be changed into another prime (110000110000553) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 55000055000076 + 55000055000077.

It is an arithmetic number, because the mean of its divisors is an integer number (55000055000077).

Almost surely, 2110000110000153 is an apocalyptic number.

It is an amenable number.

110000110000153 is a deficient number, since it is larger than the sum of its proper divisors (1).

110000110000153 is an equidigital number, since it uses as much as digits as its factorization.

110000110000153 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 15, while the sum is 13.

Adding to 110000110000153 its reverse (351000011000011), we get a palindrome (461000121000164).

The spelling of 110000110000153 in words is "one hundred ten trillion, one hundred ten million, one hundred fifty-three", and thus it is an aban number.