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1421131054351 is a prime number
BaseRepresentation
bin10100101011100010000…
…000101111110100001111
312000212012001210120121111
4110223202000233310033
5141240434022214401
63004505242312451
7204446644436341
oct24534200576417
95025161716544
101421131054351
114a8775482178
121ab511ab8727
13a4020716328
144cad6bd5291
1526e78347e51
hex14ae202fd0f

1421131054351 has 2 divisors, whose sum is σ = 1421131054352. Its totient is φ = 1421131054350.

The previous prime is 1421131054333. The next prime is 1421131054363. The reversal of 1421131054351 is 1534501311241.

It is a happy number.

It is a strong prime.

It is an emirp because it is prime and its reverse (1534501311241) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 1421131054351 - 25 = 1421131054319 is a prime.

It is a super-2 number, since 2×14211310543512 (a number of 25 digits) contains 22 as substring.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (1421131054391) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 710565527175 + 710565527176.

It is an arithmetic number, because the mean of its divisors is an integer number (710565527176).

Almost surely, 21421131054351 is an apocalyptic number.

1421131054351 is a deficient number, since it is larger than the sum of its proper divisors (1).

1421131054351 is an equidigital number, since it uses as much as digits as its factorization.

1421131054351 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 7200, while the sum is 31.

Adding to 1421131054351 its reverse (1534501311241), we get a palindrome (2955632365592).

The spelling of 1421131054351 in words is "one trillion, four hundred twenty-one billion, one hundred thirty-one million, fifty-four thousand, three hundred fifty-one".