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3254045365657 is a prime number
BaseRepresentation
bin101111010110100100001…
…101111101010110011001
3102112002021000110000000001
4233112210031331112121
5411303241103200112
610530515433213001
7454045164143602
oct57264415752631
912462230400001
103254045365657
11104503a85a029
124467a5261161
131a7b16a35388
14b36d4aaada9
15599a2794c57
hex2f5a437d599

3254045365657 has 2 divisors, whose sum is σ = 3254045365658. Its totient is φ = 3254045365656.

The previous prime is 3254045365643. The next prime is 3254045365667. The reversal of 3254045365657 is 7565635404523.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 2353588607881 + 900456757776 = 1534141^2 + 948924^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-3254045365657 is a prime.

It is a super-2 number, since 2×32540453656572 (a number of 26 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 3254045365598 and 3254045365607.

It is not a weakly prime, because it can be changed into another prime (3254045365667) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1627022682828 + 1627022682829.

It is an arithmetic number, because the mean of its divisors is an integer number (1627022682829).

Almost surely, 23254045365657 is an apocalyptic number.

It is an amenable number.

3254045365657 is a deficient number, since it is larger than the sum of its proper divisors (1).

3254045365657 is an equidigital number, since it uses as much as digits as its factorization.

3254045365657 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 45360000, while the sum is 55.

The spelling of 3254045365657 in words is "three trillion, two hundred fifty-four billion, forty-five million, three hundred sixty-five thousand, six hundred fifty-seven".