Search a number
-
+
3114401413100 = 225231144014131
BaseRepresentation
bin101101010100100000110…
…010100011011111101100
3102000201210222201012112012
4231110200302203133230
5402011243230204400
610342423010500352
7441002525136515
oct55244062433754
912021728635465
103114401413100
11aa08a045067a
124237128946b8
131978c0b66053
14aaa48881a0c
155602ce53d35
hex2d520ca37ec

3114401413100 has 18 divisors (see below), whose sum is σ = 6758251066644. Its totient is φ = 1245760565200.

The previous prime is 3114401413073. The next prime is 3114401413121. The reversal of 3114401413100 is 13141044113.

3114401413100 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

It is a super-2 number, since 2×31144014131002 (a number of 26 digits) contains 22 as substring.

It is an unprimeable number.

It is a polite number, since it can be written in 5 ways as a sum of consecutive naturals, for example, 15572006966 + ... + 15572007165.

Almost surely, 23114401413100 is an apocalyptic number.

It is an amenable number.

3114401413100 is an abundant number, since it is smaller than the sum of its proper divisors (3643849653544).

It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.

3114401413100 is a wasteful number, since it uses less digits than its factorization.

3114401413100 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 31144014145 (or 31144014138 counting only the distinct ones).

The product of its (nonzero) digits is 576, while the sum is 23.

Adding to 3114401413100 its reverse (13141044113), we get a palindrome (3127542457213).

The spelling of 3114401413100 in words is "three trillion, one hundred fourteen billion, four hundred one million, four hundred thirteen thousand, one hundred".

Divisors: 1 2 4 5 10 20 25 50 100 31144014131 62288028262 124576056524 155720070655 311440141310 622880282620 778600353275 1557200706550 3114401413100