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14125003244551 is a prime number
BaseRepresentation
bin1100110110001011101111…
…0011010000010000000111
31212000100002202120201111201
43031202323303100100013
53322411001312311201
650012534212243331
72655332514402553
oct315427363202007
955010082521451
1014125003244551
114556420103461
12170162b010547
137b5c9c864944
1436b92207c863
1519765524b801
hexcd8bbcd0407

14125003244551 has 2 divisors, whose sum is σ = 14125003244552. Its totient is φ = 14125003244550.

The previous prime is 14125003244531. The next prime is 14125003244567. The reversal of 14125003244551 is 15544230052141.

It is a strong prime.

It is a cyclic number.

It is not a de Polignac number, because 14125003244551 - 235 = 14090643506183 is a prime.

It is a super-2 number, since 2×141250032445512 (a number of 27 digits) contains 22 as substring.

It is a self number, because there is not a number n which added to its sum of digits gives 14125003244551.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (14125003244531) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 7062501622275 + 7062501622276.

It is an arithmetic number, because the mean of its divisors is an integer number (7062501622276).

Almost surely, 214125003244551 is an apocalyptic number.

14125003244551 is a deficient number, since it is larger than the sum of its proper divisors (1).

14125003244551 is an equidigital number, since it uses as much as digits as its factorization.

14125003244551 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 96000, while the sum is 37.

Adding to 14125003244551 its reverse (15544230052141), we get a palindrome (29669233296692).

The spelling of 14125003244551 in words is "fourteen trillion, one hundred twenty-five billion, three million, two hundred forty-four thousand, five hundred fifty-one".