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145012246433 is a prime number
BaseRepresentation
bin1000011100001101100…
…1110100011110100001
3111212022011010101020222
42013003121310132201
54333441113341213
6150341231412425
713322352520445
oct2070331643641
9455264111228
10145012246433
11565546a606a
1224130384115
13108a011aaa2
14703919a825
153b8ac96808
hex21c36747a1

145012246433 has 2 divisors, whose sum is σ = 145012246434. Its totient is φ = 145012246432.

The previous prime is 145012246399. The next prime is 145012246441. The reversal of 145012246433 is 334642210541.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 144557362849 + 454883584 = 380207^2 + 21328^2 .

It is a cyclic number.

It is not a de Polignac number, because 145012246433 - 218 = 145011984289 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 145012246393 and 145012246402.

It is not a weakly prime, because it can be changed into another prime (145012246333) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 72506123216 + 72506123217.

It is an arithmetic number, because the mean of its divisors is an integer number (72506123217).

Almost surely, 2145012246433 is an apocalyptic number.

It is an amenable number.

145012246433 is a deficient number, since it is larger than the sum of its proper divisors (1).

145012246433 is an equidigital number, since it uses as much as digits as its factorization.

145012246433 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 69120, while the sum is 35.

Adding to 145012246433 its reverse (334642210541), we get a palindrome (479654456974).

The spelling of 145012246433 in words is "one hundred forty-five billion, twelve million, two hundred forty-six thousand, four hundred thirty-three".