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1531386782017 is a prime number
BaseRepresentation
bin10110010010001101110…
…000111110000101000001
312102101202222101212101121
4112102031300332011001
5200042240004011032
63131302005232241
7215432122605451
oct26221560760501
95371688355347
101531386782017
11540503a08264
12208962402681
13b1541bac981
1454195d45d61
1529c7cb06a97
hex1648dc3e141

1531386782017 has 2 divisors, whose sum is σ = 1531386782018. Its totient is φ = 1531386782016.

The previous prime is 1531386781973. The next prime is 1531386782033. The reversal of 1531386782017 is 7102876831351.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 1126249440001 + 405137342016 = 1061249^2 + 636504^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-1531386782017 is a prime.

It is a super-2 number, since 2×15313867820172 (a number of 25 digits) contains 22 as substring.

It is not a weakly prime, because it can be changed into another prime (1531386782047) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 765693391008 + 765693391009.

It is an arithmetic number, because the mean of its divisors is an integer number (765693391009).

Almost surely, 21531386782017 is an apocalyptic number.

It is an amenable number.

1531386782017 is a deficient number, since it is larger than the sum of its proper divisors (1).

1531386782017 is an equidigital number, since it uses as much as digits as its factorization.

1531386782017 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 1693440, while the sum is 52.

The spelling of 1531386782017 in words is "one trillion, five hundred thirty-one billion, three hundred eighty-six million, seven hundred eighty-two thousand, seventeen".