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13421445127 is a prime number
BaseRepresentation
bin11000111111111101…
…10000000000000111
31021122100211001111021
430133332300000013
5204441342221002
610055444052011
7653411254426
oct143776600007
937570731437
1013421445127
115768068412
122726994007
13135b7b140c
149147132bd
1553844e737
hex31ffb0007

13421445127 has 2 divisors, whose sum is σ = 13421445128. Its totient is φ = 13421445126.

The previous prime is 13421445109. The next prime is 13421445203. The reversal of 13421445127 is 72154412431.

13421445127 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

It is a weak prime.

It is a cyclic number.

It is not a de Polignac number, because 13421445127 - 219 = 13420920839 is a prime.

It is a super-2 number, since 2×134214451272 (a number of 21 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 13421445092 and 13421445101.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (13421443127) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (17) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6710722563 + 6710722564.

It is an arithmetic number, because the mean of its divisors is an integer number (6710722564).

Almost surely, 213421445127 is an apocalyptic number.

13421445127 is a deficient number, since it is larger than the sum of its proper divisors (1).

13421445127 is an equidigital number, since it uses as much as digits as its factorization.

13421445127 is an odious number, because the sum of its binary digits is odd.

The product of its digits is 26880, while the sum is 34.

Adding to 13421445127 its reverse (72154412431), we get a palindrome (85575857558).

The spelling of 13421445127 in words is "thirteen billion, four hundred twenty-one million, four hundred forty-five thousand, one hundred twenty-seven".