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132220123354049 is a prime number
BaseRepresentation
bin11110000100000011100101…
…011111000001001111000001
3122100011010020201210010102112
4132010003211133001033001
5114312243303034312144
61145113030334225105
736564405625641323
oct3604034537011701
9570133221703375
10132220123354049
113914726a730a25
12129b518b955195
1358a13c5242ac3
1424916b9565613
1510445374d099e
hex7840e57c13c1

132220123354049 has 2 divisors, whose sum is σ = 132220123354050. Its totient is φ = 132220123354048.

The previous prime is 132220123354021. The next prime is 132220123354061. The reversal of 132220123354049 is 940453321022231.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 131253110730625 + 967012623424 = 11456575^2 + 983368^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-132220123354049 is a prime.

It is a super-2 number, since 2×1322201233540492 (a number of 29 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 132220123353997 and 132220123354015.

It is not a weakly prime, because it can be changed into another prime (132220123354069) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 66110061677024 + 66110061677025.

It is an arithmetic number, because the mean of its divisors is an integer number (66110061677025).

Almost surely, 2132220123354049 is an apocalyptic number.

It is an amenable number.

132220123354049 is a deficient number, since it is larger than the sum of its proper divisors (1).

132220123354049 is an equidigital number, since it uses as much as digits as its factorization.

132220123354049 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 311040, while the sum is 41.

The spelling of 132220123354049 in words is "one hundred thirty-two trillion, two hundred twenty billion, one hundred twenty-three million, three hundred fifty-four thousand, forty-nine".