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44139765049153 is a prime number
BaseRepresentation
bin10100000100101000101110…
…00001000010011101000001
312210021201110010012212000121
422002110113001002131001
521241141214323033103
6233513310024220241
712203664345263224
oct1202242701023501
9183251403185017
1044139765049153
1113078629715824
124b4a6ba00a681
131b824923815c7
14ac8539b675bb
1551829ae886bd
hex282517042741

44139765049153 has 2 divisors, whose sum is σ = 44139765049154. Its totient is φ = 44139765049152.

The previous prime is 44139765049139. The next prime is 44139765049201. The reversal of 44139765049153 is 35194056793144.

It is a happy number.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 43903186898704 + 236578150449 = 6625948^2 + 486393^2 .

It is a cyclic number.

It is not a de Polignac number, because 44139765049153 - 221 = 44139762952001 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 44139765049091 and 44139765049100.

It is not a weakly prime, because it can be changed into another prime (44139765049123) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 22069882524576 + 22069882524577.

It is an arithmetic number, because the mean of its divisors is an integer number (22069882524577).

Almost surely, 244139765049153 is an apocalyptic number.

It is an amenable number.

44139765049153 is a deficient number, since it is larger than the sum of its proper divisors (1).

44139765049153 is an equidigital number, since it uses as much as digits as its factorization.

44139765049153 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 48988800, while the sum is 61.

The spelling of 44139765049153 in words is "forty-four trillion, one hundred thirty-nine billion, seven hundred sixty-five million, forty-nine thousand, one hundred fifty-three".