Base | Representation |
---|---|
bin | 1110000110010001111011… |
… | …11100011101101111001111 |
3 | 11001202202212120111000201112 |
4 | 13003020331330131233033 |
5 | 13030414311440000101 |
6 | 145534101343002235 |
7 | 6346554160152326 |
oct | 703107574355717 |
9 | 131682776430645 |
10 | 31002113203151 |
11 | 9972a16752832 |
12 | 3588500b9937b |
13 | 143b64346944c |
14 | 7927203190bd |
15 | 38b681d146bb |
hex | 1c323df1dbcf |
31002113203151 has 2 divisors, whose sum is σ = 31002113203152. Its totient is φ = 31002113203150.
The previous prime is 31002113203129. The next prime is 31002113203163. The reversal of 31002113203151 is 15130231120013.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-31002113203151 is a prime.
It is a super-2 number, since 2×310021132031512 (a number of 28 digits) contains 22 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (31002113203051) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 15501056601575 + 15501056601576.
It is an arithmetic number, because the mean of its divisors is an integer number (15501056601576).
Almost surely, 231002113203151 is an apocalyptic number.
31002113203151 is a deficient number, since it is larger than the sum of its proper divisors (1).
31002113203151 is an equidigital number, since it uses as much as digits as its factorization.
31002113203151 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 540, while the sum is 23.
Adding to 31002113203151 its reverse (15130231120013), we get a palindrome (46132344323164).
The spelling of 31002113203151 in words is "thirty-one trillion, two billion, one hundred thirteen million, two hundred three thousand, one hundred fifty-one".
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