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1300154092169 is a prime number
BaseRepresentation
bin10010111010110111001…
…110010000101010001001
311121021220222022010221112
4102322313032100222021
5132300203421422134
62433141010220105
7162635013303602
oct22726716205211
94537828263845
101300154092169
114614350a9999
1218bb8ab84635
13957b0cc9b3a
1446cdbc1ada9
1523c477121ce
hex12eb7390a89

1300154092169 has 2 divisors, whose sum is σ = 1300154092170. Its totient is φ = 1300154092168.

The previous prime is 1300154092163. The next prime is 1300154092213. The reversal of 1300154092169 is 9612904510031.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 963999967225 + 336154124944 = 981835^2 + 579788^2 .

It is an emirp because it is prime and its reverse (9612904510031) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 1300154092169 - 212 = 1300154088073 is a prime.

It is a super-2 number, since 2×13001540921692 (a number of 25 digits) contains 22 as substring.

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 (1300154092163) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 650077046084 + 650077046085.

It is an arithmetic number, because the mean of its divisors is an integer number (650077046085).

Almost surely, 21300154092169 is an apocalyptic number.

It is an amenable number.

1300154092169 is a deficient number, since it is larger than the sum of its proper divisors (1).

1300154092169 is an equidigital number, since it uses as much as digits as its factorization.

1300154092169 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 58320, while the sum is 41.

The spelling of 1300154092169 in words is "one trillion, three hundred billion, one hundred fifty-four million, ninety-two thousand, one hundred sixty-nine".