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1031213121300 = 22325223376472081
BaseRepresentation
bin11110000000110010001…
…01110111111100010100
310122120202000102021102200
433000121011313330110
5113343411024340200
62105422140004500
7134334264622455
oct17003105677424
93576660367380
101031213121300
11368375a17a0a
12147a33230730
1376320a9a59b
1437ca7a3912c
151bc56b1d800
hexf019177f14

1031213121300 has 432 divisors, whose sum is σ = 3470992342272. Its totient is φ = 255406694400.

The previous prime is 1031213121287. The next prime is 1031213121347. The reversal of 1031213121300 is 31213121301.

1031213121300 is a `hidden beast` number, since 1 + 0 + 31 + 21 + 312 + 1 + 300 = 666.

1031213121300 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

It is a Harshad number since it is a multiple of its sum of digits (18).

It is a congruent number.

It is an unprimeable number.

It is a polite number, since it can be written in 143 ways as a sum of consecutive naturals, for example, 495536260 + ... + 495538340.

It is an arithmetic number, because the mean of its divisors is an integer number (8034704496).

Almost surely, 21031213121300 is an apocalyptic number.

1031213121300 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 1031213121300, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (1735496171136).

1031213121300 is an abundant number, since it is smaller than the sum of its proper divisors (2439779220972).

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

1031213121300 is a wasteful number, since it uses less digits than its factorization.

1031213121300 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 2808 (or 2798 counting only the distinct ones).

The product of its (nonzero) digits is 108, while the sum is 18.

Adding to 1031213121300 its reverse (31213121301), we get a palindrome (1062426242601).

Subtracting from 1031213121300 its reverse (31213121301), we obtain a palindrome (999999999999).

The spelling of 1031213121300 in words is "one trillion, thirty-one billion, two hundred thirteen million, one hundred twenty-one thousand, three hundred".