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1303974945341 is a prime number
BaseRepresentation
bin10010111110011010111…
…101101010001000111101
311121122210020220121010222
4102332122331222020331
5132331020041222331
62435012100311125
7163131500104325
oct22763275521075
94548706817128
101303974945341
11463015953492
121908766a24a5
1395c6c785ac3
144718145a285
1523dbcd9bb7b
hex12f9af6a23d

1303974945341 has 2 divisors, whose sum is σ = 1303974945342. Its totient is φ = 1303974945340.

The previous prime is 1303974945239. The next prime is 1303974945373. The reversal of 1303974945341 is 1435494793031.

It is a happy number.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 958059228025 + 345915717316 = 978805^2 + 588146^2 .

It is a cyclic number.

It is not a de Polignac number, because 1303974945341 - 210 = 1303974944317 is a prime.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (1303974945391) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 651987472670 + 651987472671.

It is an arithmetic number, because the mean of its divisors is an integer number (651987472671).

Almost surely, 21303974945341 is an apocalyptic number.

It is an amenable number.

1303974945341 is a deficient number, since it is larger than the sum of its proper divisors (1).

1303974945341 is an equidigital number, since it uses as much as digits as its factorization.

1303974945341 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 4898880, while the sum is 53.

The spelling of 1303974945341 in words is "one trillion, three hundred three billion, nine hundred seventy-four million, nine hundred forty-five thousand, three hundred forty-one".