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32232555011441 is a prime number
BaseRepresentation
bin1110101010000101110011…
…11111011101110101110001
311020010101212011211020010002
413111002321333131311301
513211044233040331231
6152315233101115345
76534503522025131
oct725027177356561
9136111764736102
1032232555011441
11a2a7828003a74
123746a75558b55
1314ca694808465
147d60c7816ac1
153ad6992ea0cb
hex1d50b9fddd71

32232555011441 has 2 divisors, whose sum is σ = 32232555011442. Its totient is φ = 32232555011440.

The previous prime is 32232555011411. The next prime is 32232555011443. The reversal of 32232555011441 is 14411055523223.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 28169927776225 + 4062627235216 = 5307535^2 + 2015596^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-32232555011441 is a prime.

It is a Sophie Germain prime.

Together with 32232555011443, it forms a pair of twin primes.

It is a Chen prime.

It is a Curzon number.

It is a junction number, because it is equal to n+sod(n) for n = 32232555011395 and 32232555011404.

It is not a weakly prime, because it can be changed into another prime (32232555011443) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 16116277505720 + 16116277505721.

It is an arithmetic number, because the mean of its divisors is an integer number (16116277505721).

Almost surely, 232232555011441 is an apocalyptic number.

It is an amenable number.

32232555011441 is a deficient number, since it is larger than the sum of its proper divisors (1).

32232555011441 is an equidigital number, since it uses as much as digits as its factorization.

32232555011441 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 144000, while the sum is 38.

The spelling of 32232555011441 in words is "thirty-two trillion, two hundred thirty-two billion, five hundred fifty-five million, eleven thousand, four hundred forty-one".