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20131300021320 = 23325712293439343
BaseRepresentation
bin1001001001111001011101…
…11111001001010001001000
32122021112101220121200021100
410210330232333021101020
510114312400301140240
6110452105213141400
74145303244545565
oct444745677112110
978245356550240
1020131300021320
116461700049454
1223116b778a860
13b304b1600147
144d850825186c
1524d9dbde0e30
hex124f2efc9448

20131300021320 has 192 divisors, whose sum is σ = 66637977868800. Its totient is φ = 5269622238720.

The previous prime is 20131300021279. The next prime is 20131300021331. The reversal of 20131300021320 is 2312000313102.

20131300021320 is a `hidden beast` number, since 2 + 0 + 131 + 300 + 0 + 213 + 20 = 666.

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

It is a junction number, because it is equal to n+sod(n) for n = 20131300021293 and 20131300021302.

It is an unprimeable number.

It is a polite number, since it can be written in 47 ways as a sum of consecutive naturals, for example, 4133569 + ... + 7572911.

It is an arithmetic number, because the mean of its divisors is an integer number (347072801400).

Almost surely, 220131300021320 is an apocalyptic number.

20131300021320 is a gapful number since it is divisible by the number (20) 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 20131300021320, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (33318988934400).

20131300021320 is an abundant number, since it is smaller than the sum of its proper divisors (46506677847480).

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

20131300021320 is a wasteful number, since it uses less digits than its factorization.

20131300021320 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 3439660 (or 3439653 counting only the distinct ones).

The product of its (nonzero) digits is 216, while the sum is 18.

Adding to 20131300021320 its reverse (2312000313102), we get a palindrome (22443300334422).

The spelling of 20131300021320 in words is "twenty trillion, one hundred thirty-one billion, three hundred million, twenty-one thousand, three hundred twenty".