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12001400313433 = 473947527312269
BaseRepresentation
bin1010111010100100101100…
…1100111101011001011001
31120111022201020000101200221
42232221023030331121121
53033112331440012213
641305211151400041
72346033613331656
oct256511314753131
946438636011627
1012001400313433
113907846668033
121419b50823021
1369095b84ac0b
142d6c29972c2d
1515c2b5d2838d
hexaea4b33d659

12001400313433 has 16 divisors (see below), whose sum is σ = 12263179057920. Its totient is φ = 11739891454336.

The previous prime is 12001400313403. The next prime is 12001400313443. The reversal of 12001400313433 is 33431300410021.

It is a cyclic number.

It is not a de Polignac number, because 12001400313433 - 213 = 12001400305241 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 12001400313398 and 12001400313407.

It is not an unprimeable number, because it can be changed into a prime (12001400313403) by changing a digit.

It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 978182823 + ... + 978195091.

It is an arithmetic number, because the mean of its divisors is an integer number (766448691120).

Almost surely, 212001400313433 is an apocalyptic number.

It is an amenable number.

12001400313433 is a deficient number, since it is larger than the sum of its proper divisors (261778744487).

12001400313433 is a wasteful number, since it uses less digits than its factorization.

12001400313433 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 21536.

The product of its (nonzero) digits is 2592, while the sum is 25.

Adding to 12001400313433 its reverse (33431300410021), we get a palindrome (45432700723454).

The spelling of 12001400313433 in words is "twelve trillion, one billion, four hundred million, three hundred thirteen thousand, four hundred thirty-three".

Divisors: 1 47 3947 5273 12269 185509 247831 576643 20812531 48425743 64694437 978188957 2276009921 3040638539 255348942839 12001400313433