Base | Representation |
---|---|
bin | 10111001111110111111101… |
… | …00000110111000110010011 |
3 | 20201000022022201210202211211 |
4 | 23213323332200313012103 |
5 | 23200044321010043042 |
6 | 300421322450320551 |
7 | 13524340605533623 |
oct | 1347737640670623 |
9 | 221008281722754 |
10 | 51122970784147 |
11 | 15320148264209 |
12 | 5897b83589157 |
13 | 226ab4c4cb302 |
14 | c8a517756883 |
15 | 5d9c5c09e017 |
hex | 2e7efe837193 |
51122970784147 has 2 divisors, whose sum is σ = 51122970784148. Its totient is φ = 51122970784146.
The previous prime is 51122970784111. The next prime is 51122970784157. The reversal of 51122970784147 is 74148707922115.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 51122970784147 - 223 = 51122962395539 is a prime.
It is a super-2 number, since 2×511229707841472 (a number of 28 digits) contains 22 as substring.
It is a junction number, because it is equal to n+sod(n) for n = 51122970784091 and 51122970784100.
It is not a weakly prime, because it can be changed into another prime (51122970784157) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 25561485392073 + 25561485392074.
It is an arithmetic number, because the mean of its divisors is an integer number (25561485392074).
Almost surely, 251122970784147 is an apocalyptic number.
51122970784147 is a deficient number, since it is larger than the sum of its proper divisors (1).
51122970784147 is an equidigital number, since it uses as much as digits as its factorization.
51122970784147 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 7902720, while the sum is 58.
The spelling of 51122970784147 in words is "fifty-one trillion, one hundred twenty-two billion, nine hundred seventy million, seven hundred eighty-four thousand, one hundred forty-seven".
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