Base | Representation |
---|---|
bin | 10100010011011110111… |
… | …111001010000001101101 |
3 | 11221101112120212121120122 |
4 | 110103132333022001231 |
5 | 140330043442030442 |
6 | 2544555150254325 |
7 | 202544044402136 |
oct | 24233677120155 |
9 | 4841476777518 |
10 | 1395310501997 |
11 | 498825448005 |
12 | 1a65067b73a5 |
13 | a17671b2299 |
14 | 4b767892c8d |
15 | 264665dddd2 |
hex | 144defca06d |
1395310501997 has 4 divisors (see below), whose sum is σ = 1395313338816. Its totient is φ = 1395307665180.
The previous prime is 1395310501993. The next prime is 1395310502021. The reversal of 1395310501997 is 7991050135931.
It is a semiprime because it is the product of two primes, and also a Blum integer, because the two primes are equal to 3 mod 4.
It is a cyclic number.
It is not a de Polignac number, because 1395310501997 - 22 = 1395310501993 is a prime.
It is a super-4 number, since 4×13953105019974 (a number of 50 digits) contains 4444 as substring.
It is a Duffinian number.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (1395310501993) by changing a digit.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 468629 + ... + 1735002.
It is an arithmetic number, because the mean of its divisors is an integer number (348828334704).
Almost surely, 21395310501997 is an apocalyptic number.
It is an amenable number.
1395310501997 is a deficient number, since it is larger than the sum of its proper divisors (2836819).
1395310501997 is an equidigital number, since it uses as much as digits as its factorization.
1395310501997 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 2836818.
The product of its (nonzero) digits is 1148175, while the sum is 53.
The spelling of 1395310501997 in words is "one trillion, three hundred ninety-five billion, three hundred ten million, five hundred one thousand, nine hundred ninety-seven".
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