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11403132000000 = 2835613671091
BaseRepresentation
bin1010010111101111111110…
…1000011101011100000000
31111101010110212122010121020
42211323333220131130000
52443312103243000000
640130305421225440
72254564145510556
oct245737750353400
944333425563536
1011403132000000
1136a704a80a264
121342005a02280
136494068b2820
142b5cb36a1bd6
1514b94d01cba0
hexa5effa1d700

11403132000000 has 1008 divisors, whose sum is σ = 41501611272576. Its totient is φ = 2762496000000.

The previous prime is 11403131999969. The next prime is 11403132000049. The reversal of 11403132000000 is 23130411.

It is a Harshad number since it is a multiple of its sum of digits (15).

It is a congruent number.

It is an unprimeable number.

It is a polite number, since it can be written in 111 ways as a sum of consecutive naturals, for example, 10451999455 + ... + 10452000545.

Almost surely, 211403132000000 is an apocalyptic number.

11403132000000 is a gapful number since it is divisible by the number (10) formed by its first and last digit.

It is an amenable number.

It is a practical number, because each smaller number is the sum of distinct divisors of 11403132000000, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (20750805636288).

11403132000000 is an abundant number, since it is smaller than the sum of its proper divisors (30098479272576).

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

11403132000000 is an frugal number, since it uses more digits than its factorization.

11403132000000 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 1220 (or 1181 counting only the distinct ones).

The product of its (nonzero) digits is 72, while the sum is 15.

Adding to 11403132000000 its reverse (23130411), we get a palindrome (11403155130411).

The spelling of 11403132000000 in words is "eleven trillion, four hundred three billion, one hundred thirty-two million", and thus it is an aban number.