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130043767641 = 343347922547
BaseRepresentation
bin111100100011100110…
…1100110101101011001
3110102122222021212112120
41321013031212231121
54112312201031031
6135424041131453
712252414533544
oct1710715465531
9412588255476
10130043767641
1150173353437
1221253489b89
13c355c8a248
14641921d85b
1535b1b0ba96
hex1e47366b59

130043767641 has 4 divisors (see below), whose sum is σ = 173391690192. Its totient is φ = 86695845092.

The previous prime is 130043767627. The next prime is 130043767663. The reversal of 130043767641 is 146767340031.

It is a semiprime because it is the product of two primes, and also a Blum integer, because the two primes are equal to 3 mod 4.

It is a cyclic number.

It is not a de Polignac number, because 130043767641 - 29 = 130043767129 is a prime.

It is a super-2 number, since 2×1300437676412 (a number of 23 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 130043767593 and 130043767602.

It is not an unprimeable number, because it can be changed into a prime (130043767141) by changing a digit.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 21673961271 + ... + 21673961276.

It is an arithmetic number, because the mean of its divisors is an integer number (43347922548).

Almost surely, 2130043767641 is an apocalyptic number.

It is an amenable number.

130043767641 is a deficient number, since it is larger than the sum of its proper divisors (43347922551).

130043767641 is an equidigital number, since it uses as much as digits as its factorization.

130043767641 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 43347922550.

The product of its (nonzero) digits is 254016, while the sum is 42.

The spelling of 130043767641 in words is "one hundred thirty billion, forty-three million, seven hundred sixty-seven thousand, six hundred forty-one".

Divisors: 1 3 43347922547 130043767641