Base | Representation |
---|---|
bin | 10001101100000100100… |
… | …010111010001111111001 |
3 | 11022012112220112211022211 |
4 | 101230010202322033321 |
5 | 124403422303130423 |
6 | 2330230014015121 |
7 | 153551340364555 |
oct | 21540442721771 |
9 | 4265486484284 |
10 | 1215552005113 |
11 | 429570465786 |
12 | 1776b9aa74a1 |
13 | 8a81a5c6ab4 |
14 | 42b93ba0865 |
15 | 219451dc10d |
hex | 11b048ba3f9 |
1215552005113 has 2 divisors, whose sum is σ = 1215552005114. Its totient is φ = 1215552005112.
The previous prime is 1215552005021. The next prime is 1215552005119. The reversal of 1215552005113 is 3115002555121.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 946801003369 + 268751001744 = 973037^2 + 518412^2 .
It is a cyclic number.
It is not a de Polignac number, because 1215552005113 - 217 = 1215551874041 is a prime.
It is not a weakly prime, because it can be changed into another prime (1215552005119) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 607776002556 + 607776002557.
It is an arithmetic number, because the mean of its divisors is an integer number (607776002557).
Almost surely, 21215552005113 is an apocalyptic number.
It is an amenable number.
1215552005113 is a deficient number, since it is larger than the sum of its proper divisors (1).
1215552005113 is an equidigital number, since it uses as much as digits as its factorization.
1215552005113 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 7500, while the sum is 31.
The spelling of 1215552005113 in words is "one trillion, two hundred fifteen billion, five hundred fifty-two million, five thousand, one hundred thirteen".
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