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11403034121434 = 25701517060717
BaseRepresentation
bin1010010111101111100111…
…0011000101010011011010
31111101010020001110101110011
42211323321303011103122
52443311403213341214
640130252003312134
72254561546532404
oct245737163052332
944333201411404
1011403034121434
1136a6aaa53768a
121341b9907b64a
136493bc532807
142b5ca46a3b74
1514b944636ac4
hexa5ef9cc54da

11403034121434 has 4 divisors (see below), whose sum is σ = 17104551182154. Its totient is φ = 5701517060716.

The previous prime is 11403034121399. The next prime is 11403034121447. The reversal of 11403034121434 is 43412143030411.

It is a semiprime because it is the product of two primes.

It can be written as a sum of positive squares in only one way, i.e., 10265712120225 + 1137322001209 = 3204015^2 + 1066453^2 .

It is a junction number, because it is equal to n+sod(n) for n = 11403034121396 and 11403034121405.

It is an unprimeable number.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2850758530357 + ... + 2850758530360.

Almost surely, 211403034121434 is an apocalyptic number.

11403034121434 is a deficient number, since it is larger than the sum of its proper divisors (5701517060720).

11403034121434 is an equidigital number, since it uses as much as digits as its factorization.

11403034121434 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 5701517060719.

The product of its (nonzero) digits is 13824, while the sum is 31.

Adding to 11403034121434 its reverse (43412143030411), we get a palindrome (54815177151845).

The spelling of 11403034121434 in words is "eleven trillion, four hundred three billion, thirty-four million, one hundred twenty-one thousand, four hundred thirty-four".

Divisors: 1 2 5701517060717 11403034121434