Search a number
-
+
311411054434 = 2155705527217
BaseRepresentation
bin1001000100000011000…
…10111001011101100010
31002202210202211122122211
410202001202321131202
520100232212220214
6355021003352334
731333025200534
oct4420142713542
91082722748584
10311411054434
111100834703a7
125042ab226aa
132349ac6831c
1411102853854
1581793970c4
hex48818b9762

311411054434 has 4 divisors (see below), whose sum is σ = 467116581654. Its totient is φ = 155705527216.

The previous prime is 311411054413. The next prime is 311411054453. The reversal of 311411054434 is 434450114113.

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., 296355028225 + 15056026209 = 544385^2 + 122703^2 .

It is a super-2 number, since 2×3114110544342 (a number of 24 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 311411054396 and 311411054405.

It is an unprimeable number.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 77852763607 + ... + 77852763610.

Almost surely, 2311411054434 is an apocalyptic number.

311411054434 is a deficient number, since it is larger than the sum of its proper divisors (155705527220).

311411054434 is a wasteful number, since it uses less digits than its factorization.

311411054434 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 155705527219.

The product of its (nonzero) digits is 11520, while the sum is 31.

Adding to 311411054434 its reverse (434450114113), we get a palindrome (745861168547).

The spelling of 311411054434 in words is "three hundred eleven billion, four hundred eleven million, fifty-four thousand, four hundred thirty-four".

Divisors: 1 2 155705527217 311411054434