Base | Representation |
---|---|
bin | 1100001010111… |
… | …1101001101001 |
3 | 10120001122221222 |
4 | 3002233221221 |
5 | 101032113423 |
6 | 5022111425 |
7 | 1156633031 |
oct | 302575151 |
9 | 116048858 |
10 | 51051113 |
11 | 268a9503 |
12 | 1511b575 |
13 | a765979 |
14 | 6acc8c1 |
15 | 47363c8 |
hex | 30afa69 |
51051113 has 2 divisors, whose sum is σ = 51051114. Its totient is φ = 51051112.
The previous prime is 51051103. The next prime is 51051127. The reversal of 51051113 is 31115015.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 50651689 + 399424 = 7117^2 + 632^2 .
It is a cyclic number.
It is not a de Polignac number, because 51051113 - 24 = 51051097 is a prime.
It is a super-2 number, since 2×510511132 = 5212432277077538, which contains 22 as substring.
It is a Chen prime.
It is a junction number, because it is equal to n+sod(n) for n = 51051091 and 51051100.
It is not a weakly prime, because it can be changed into another prime (51051103) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 25525556 + 25525557.
It is an arithmetic number, because the mean of its divisors is an integer number (25525557).
Almost surely, 251051113 is an apocalyptic number.
It is an amenable number.
51051113 is a deficient number, since it is larger than the sum of its proper divisors (1).
51051113 is an equidigital number, since it uses as much as digits as its factorization.
51051113 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 75, while the sum is 17.
The square root of 51051113 is about 7145.0061581499. The cubic root of 51051113 is about 370.9668237783.
Adding to 51051113 its reverse (31115015), we get a palindrome (82166128).
It can be divided in two parts, 5105 and 1113, that multiplied together give a palindrome (5681865).
The spelling of 51051113 in words is "fifty-one million, fifty-one thousand, one hundred thirteen".
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