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144034331297653 is a prime number
BaseRepresentation
bin100000101111111110011011…
…001001110000011101110101
3200212222112210012022212022001
4200233332123021300131311
5122334324302303011103
61230200242555535301
742224065000356046
oct4057763311603565
9625875705285261
10144034331297653
1141991687060985
12141a299a15a531
13624a4c102b234
14277d427dc88cd
15119b9e448801d
hex82ff9b270775

144034331297653 has 2 divisors, whose sum is σ = 144034331297654. Its totient is φ = 144034331297652.

The previous prime is 144034331297651. The next prime is 144034331297663. The reversal of 144034331297653 is 356792133430441.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 111924508395844 + 32109822901809 = 10579438^2 + 5666553^2 .

It is a cyclic number.

It is not a de Polignac number, because 144034331297653 - 21 = 144034331297651 is a prime.

Together with 144034331297651, it forms a pair of twin primes.

It is a junction number, because it is equal to n+sod(n) for n = 144034331297594 and 144034331297603.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (144034331297651) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 72017165648826 + 72017165648827.

It is an arithmetic number, because the mean of its divisors is an integer number (72017165648827).

Almost surely, 2144034331297653 is an apocalyptic number.

It is an amenable number.

144034331297653 is a deficient number, since it is larger than the sum of its proper divisors (1).

144034331297653 is an equidigital number, since it uses as much as digits as its factorization.

144034331297653 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 19595520, while the sum is 55.

The spelling of 144034331297653 in words is "one hundred forty-four trillion, thirty-four billion, three hundred thirty-one million, two hundred ninety-seven thousand, six hundred fifty-three".