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3151132021051 = 7450161717293
BaseRepresentation
bin101101110110101110000…
…110101101000100111011
3102011020121210211102021021
4231312232012231010323
5403112004244133201
610411335435123311
7443442663145210
oct55665606550473
912136553742237
103151132021051
111005429970013
1242a8638a3537
1319b1c5791bb1
14ac730b5db07
1556e7c8d1ea1
hex2ddae1ad13b

3151132021051 has 4 divisors (see below), whose sum is σ = 3601293738352. Its totient is φ = 2700970303752.

The previous prime is 3151132021049. The next prime is 3151132021067. The reversal of 3151132021051 is 1501202311513.

It is a semiprime because it is the product of two primes.

It is not a de Polignac number, because 3151132021051 - 21 = 3151132021049 is a prime.

It is a Duffinian number.

It is not an unprimeable number, because it can be changed into a prime (3151132025051) by changing a digit.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 225080858640 + ... + 225080858653.

It is an arithmetic number, because the mean of its divisors is an integer number (900323434588).

Almost surely, 23151132021051 is an apocalyptic number.

3151132021051 is a deficient number, since it is larger than the sum of its proper divisors (450161717301).

3151132021051 is an equidigital number, since it uses as much as digits as its factorization.

3151132021051 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 450161717300.

The product of its (nonzero) digits is 900, while the sum is 25.

Adding to 3151132021051 its reverse (1501202311513), we get a palindrome (4652334332564).

The spelling of 3151132021051 in words is "three trillion, one hundred fifty-one billion, one hundred thirty-two million, twenty-one thousand, fifty-one".

Divisors: 1 7 450161717293 3151132021051