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2231209943447 is a prime number
BaseRepresentation
bin100000011101111110011…
…110011110000110010111
321220022010212010221012212
4200131332132132012113
5243024004221142242
64425000444304035
7320125314411602
oct40357636360627
97808125127185
102231209943447
11790283933481
1230050b84201b
1313252c8119b4
1479dc3977339
153d08b3cc782
hex2077e79e197

2231209943447 has 2 divisors, whose sum is σ = 2231209943448. Its totient is φ = 2231209943446.

The previous prime is 2231209943399. The next prime is 2231209943489. The reversal of 2231209943447 is 7443499021322.

It is a strong prime.

It is a cyclic number.

It is not a de Polignac number, because 2231209943447 - 218 = 2231209681303 is a prime.

It is a super-3 number, since 3×22312099434473 (a number of 38 digits) contains 333 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 2231209943395 and 2231209943404.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (2231209943497) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1115604971723 + 1115604971724.

It is an arithmetic number, because the mean of its divisors is an integer number (1115604971724).

Almost surely, 22231209943447 is an apocalyptic number.

2231209943447 is a deficient number, since it is larger than the sum of its proper divisors (1).

2231209943447 is an equidigital number, since it uses as much as digits as its factorization.

2231209943447 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 2612736, while the sum is 50.

The spelling of 2231209943447 in words is "two trillion, two hundred thirty-one billion, two hundred nine million, nine hundred forty-three thousand, four hundred forty-seven".