Base | Representation |
---|---|
bin | 10111010000000110111101… |
… | …01001011111100100011111 |
3 | 20201001002010010022202120121 |
4 | 23220003132221133210133 |
5 | 23200212320141021001 |
6 | 300425134044055411 |
7 | 13525044330635056 |
oct | 1350033651374437 |
9 | 221032103282517 |
10 | 51131026110751 |
11 | 15323601290324 |
12 | 5899651234567 |
13 | 226b834343162 |
14 | c8aa7d50339d |
15 | 5da07e3776a1 |
hex | 2e80dea5f91f |
51131026110751 has 2 divisors, whose sum is σ = 51131026110752. Its totient is φ = 51131026110750.
The previous prime is 51131026110727. The next prime is 51131026110757. The reversal of 51131026110751 is 15701162013115.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 51131026110751 - 229 = 51130489239839 is a prime.
It is a super-2 number, since 2×511310261107512 (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 (51131026110757) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 25565513055375 + 25565513055376.
It is an arithmetic number, because the mean of its divisors is an integer number (25565513055376).
Almost surely, 251131026110751 is an apocalyptic number.
51131026110751 is a deficient number, since it is larger than the sum of its proper divisors (1).
51131026110751 is an equidigital number, since it uses as much as digits as its factorization.
51131026110751 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 6300, while the sum is 34.
Adding to 51131026110751 its reverse (15701162013115), we get a palindrome (66832188123866).
The spelling of 51131026110751 in words is "fifty-one trillion, one hundred thirty-one billion, twenty-six million, one hundred ten thousand, seven hundred fifty-one".
• e-mail: info -at- numbersaplenty.com • Privacy notice • done in 0.074 sec. • engine limits •