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132017935649 is a prime number
BaseRepresentation
bin111101011110011100…
…0011101010100100001
3110121202120000222221012
41322330320131110201
54130333042420044
6140352002310305
712352345636445
oct1727470352441
9417676028835
10132017935649
1150a96757046
1221704658395
13c5abc8b7b2
1465654b2825
15367a0ba59e
hex1ebce1d521

132017935649 has 2 divisors, whose sum is σ = 132017935650. Its totient is φ = 132017935648.

The previous prime is 132017935627. The next prime is 132017935739. The reversal of 132017935649 is 946539710231.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 71285262049 + 60732673600 = 266993^2 + 246440^2 .

It is an emirp because it is prime and its reverse (946539710231) is a distict prime.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-132017935649 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 132017935597 and 132017935606.

It is not a weakly prime, because it can be changed into another prime (132017935949) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 66008967824 + 66008967825.

It is an arithmetic number, because the mean of its divisors is an integer number (66008967825).

Almost surely, 2132017935649 is an apocalyptic number.

It is an amenable number.

132017935649 is a deficient number, since it is larger than the sum of its proper divisors (1).

132017935649 is an equidigital number, since it uses as much as digits as its factorization.

132017935649 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 1224720, while the sum is 50.

The spelling of 132017935649 in words is "one hundred thirty-two billion, seventeen million, nine hundred thirty-five thousand, six hundred forty-nine".