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122554311004393 is a prime number
BaseRepresentation
bin11011110111011001100110…
…001011000101000011101001
3121001221002002122120111001021
4123313121212023011003221
5112030412212104120033
61112352401230252441
734546154034402352
oct3367314613050351
9531832078514037
10122554311004393
113605aa9a639994
12118b3a1a957121
13534ca8ba51aa8
142239935c22d29
15e27dbb315b2d
hex6f76662c50e9

122554311004393 has 2 divisors, whose sum is σ = 122554311004394. Its totient is φ = 122554311004392.

The previous prime is 122554311004361. The next prime is 122554311004421. The reversal of 122554311004393 is 393400113455221.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 122249885148889 + 304425855504 = 11056667^2 + 551748^2 .

It is a cyclic number.

It is not a de Polignac number, because 122554311004393 - 25 = 122554311004361 is a prime.

It is a super-2 number, since 2×1225543110043932 (a number of 29 digits) contains 22 as substring.

It is not a weakly prime, because it can be changed into another prime (122554311004093) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 61277155502196 + 61277155502197.

It is an arithmetic number, because the mean of its divisors is an integer number (61277155502197).

Almost surely, 2122554311004393 is an apocalyptic number.

It is an amenable number.

122554311004393 is a deficient number, since it is larger than the sum of its proper divisors (1).

122554311004393 is an equidigital number, since it uses as much as digits as its factorization.

122554311004393 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 388800, while the sum is 43.

The spelling of 122554311004393 in words is "one hundred twenty-two trillion, five hundred fifty-four billion, three hundred eleven million, four thousand, three hundred ninety-three".