Base | Representation |
---|---|
bin | 10010101011110100… |
… | …11011000011111111 |
3 | 221220002200111200201 |
4 | 21111322123003333 |
5 | 131021004010101 |
6 | 4335221504331 |
7 | 503404516636 |
oct | 112572330377 |
9 | 27802614621 |
10 | 10031313151 |
11 | 4288462052 |
12 | 1b3b5720a7 |
13 | c3b332072 |
14 | 6b239451d |
15 | 3da9e5e01 |
hex | 255e9b0ff |
10031313151 has 2 divisors, whose sum is σ = 10031313152. Its totient is φ = 10031313150.
The previous prime is 10031313107. The next prime is 10031313161. The reversal of 10031313151 is 15131313001.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-10031313151 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (10031313161) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5015656575 + 5015656576.
It is an arithmetic number, because the mean of its divisors is an integer number (5015656576).
Almost surely, 210031313151 is an apocalyptic number.
10031313151 is a deficient number, since it is larger than the sum of its proper divisors (1).
10031313151 is an equidigital number, since it uses as much as digits as its factorization.
10031313151 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 135, while the sum is 19.
Adding to 10031313151 its reverse (15131313001), we get a palindrome (25162626152).
The spelling of 10031313151 in words is "ten billion, thirty-one million, three hundred thirteen thousand, one hundred fifty-one".
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