Search a number
-
+
13000000004022 = 2319751109699087
BaseRepresentation
bin1011110100101100110001…
…1000011101111110110110
31201000210021100111201210110
42331023030120131332312
53200443000000112042
643352041220403450
72511135055546203
oct275131430357666
951023240451713
1013000000004022
1141622a6102819
12155b5a1846586
13733b8298a1b5
1432d2bc026caa
1517825e68809c
hexbd2cc61dfb6

13000000004022 has 16 divisors (see below), whose sum is σ = 26001316634112. Its totient is φ = 4333113897000.

The previous prime is 13000000004011. The next prime is 13000000004023. The reversal of 13000000004022 is 22040000000031.

It is a junction number, because it is equal to n+sod(n) for n = 13000000003989 and 13000000004007.

It is a congruent number.

It is not an unprimeable number, because it can be changed into a prime (13000000004023) by changing a digit.

It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 54731038 + ... + 54968049.

It is an arithmetic number, because the mean of its divisors is an integer number (1625082289632).

Almost surely, 213000000004022 is an apocalyptic number.

13000000004022 is an abundant number, since it is smaller than the sum of its proper divisors (13001316630090).

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

13000000004022 is a wasteful number, since it uses less digits than its factorization.

13000000004022 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 109718843.

The product of its (nonzero) digits is 48, while the sum is 12.

Adding to 13000000004022 its reverse (22040000000031), we get a palindrome (35040000004053).

The spelling of 13000000004022 in words is "thirteen trillion, four thousand, twenty-two".

Divisors: 1 2 3 6 19751 39502 59253 118506 109699087 219398174 329097261 658194522 2166666667337 4333333334674 6500000002011 13000000004022