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105515431433 is a prime number
BaseRepresentation
bin110001001000100110…
…1011011011000001001
3101002100112220021210222
41202101031123120021
53212043422301213
6120250105515425
710423525240346
oct1422115333011
9332315807728
10105515431433
114082678a3a5
1218548aa2b75
139c47397239
14516d7737cd
152b28556808
hex189135b609

105515431433 has 2 divisors, whose sum is σ = 105515431434. Its totient is φ = 105515431432.

The previous prime is 105515431393. The next prime is 105515431499. The reversal of 105515431433 is 334134515501.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 100709753104 + 4805678329 = 317348^2 + 69323^2 .

It is a cyclic number.

It is not a de Polignac number, because 105515431433 - 230 = 104441689609 is a prime.

It is a super-2 number, since 2×1055154314332 (a number of 23 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 105515431393 and 105515431402.

It is not a weakly prime, because it can be changed into another prime (105515431033) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 52757715716 + 52757715717.

It is an arithmetic number, because the mean of its divisors is an integer number (52757715717).

Almost surely, 2105515431433 is an apocalyptic number.

It is an amenable number.

105515431433 is a deficient number, since it is larger than the sum of its proper divisors (1).

105515431433 is an equidigital number, since it uses as much as digits as its factorization.

105515431433 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 54000, while the sum is 35.

Adding to 105515431433 its reverse (334134515501), we get a palindrome (439649946934).

The spelling of 105515431433 in words is "one hundred five billion, five hundred fifteen million, four hundred thirty-one thousand, four hundred thirty-three".