Base | Representation |
---|---|
bin | 100110100101001100000… |
… | …1010000010011000100101 |
3 | 200202220210022001000210211 |
4 | 1031022120022002120211 |
5 | 1143334101324130201 |
6 | 15135542050215421 |
7 | 1055044641553402 |
oct | 115123012023045 |
9 | 20686708030724 |
10 | 5302542411301 |
11 | 1764882a55023 |
12 | 717802884571 |
13 | 2c6048851555 |
14 | 1449044192a9 |
15 | 92de81d7651 |
hex | 4d298282625 |
5302542411301 has 2 divisors, whose sum is σ = 5302542411302. Its totient is φ = 5302542411300.
The previous prime is 5302542411253. The next prime is 5302542411307. The reversal of 5302542411301 is 1031142452035.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 3956538701025 + 1346003710276 = 1989105^2 + 1160174^2 .
It is a cyclic number.
It is not a de Polignac number, because 5302542411301 - 227 = 5302408193573 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (5302542411307) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2651271205650 + 2651271205651.
It is an arithmetic number, because the mean of its divisors is an integer number (2651271205651).
Almost surely, 25302542411301 is an apocalyptic number.
It is an amenable number.
5302542411301 is a deficient number, since it is larger than the sum of its proper divisors (1).
5302542411301 is an equidigital number, since it uses as much as digits as its factorization.
5302542411301 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 14400, while the sum is 31.
Adding to 5302542411301 its reverse (1031142452035), we get a palindrome (6333684863336).
The spelling of 5302542411301 in words is "five trillion, three hundred two billion, five hundred forty-two million, four hundred eleven thousand, three hundred one".
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