Base | Representation |
---|---|
bin | 10111001111001101011011… |
… | …01110010011000001001111 |
3 | 20200221010101122121221202221 |
4 | 23213212231232103001033 |
5 | 23144211041432443112 |
6 | 300403025214400211 |
7 | 13522602055463434 |
oct | 1347465556230117 |
9 | 220833348557687 |
10 | 51100140515407 |
11 | 153114a2163415 |
12 | 5893671800067 |
13 | 2268951659059 |
14 | c8938d60bb8b |
15 | 5d9372b19e07 |
hex | 2e79adb9304f |
51100140515407 has 2 divisors, whose sum is σ = 51100140515408. Its totient is φ = 51100140515406.
The previous prime is 51100140515387. The next prime is 51100140515453. The reversal of 51100140515407 is 70451504100115.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 51100140515407 - 215 = 51100140482639 is a prime.
It is a super-3 number, since 3×511001405154073 (a number of 42 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (51100140515477) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 25550070257703 + 25550070257704.
It is an arithmetic number, because the mean of its divisors is an integer number (25550070257704).
Almost surely, 251100140515407 is an apocalyptic number.
51100140515407 is a deficient number, since it is larger than the sum of its proper divisors (1).
51100140515407 is an equidigital number, since it uses as much as digits as its factorization.
51100140515407 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 14000, while the sum is 34.
The spelling of 51100140515407 in words is "fifty-one trillion, one hundred billion, one hundred forty million, five hundred fifteen thousand, four hundred seven".
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