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31150500504133 is a prime number
BaseRepresentation
bin1110001010100110010101…
…00001000110001001000101
311002021221212221100221102211
413011103022201012021011
513040332211112113013
6150130201541503421
76363356304333223
oct705231241061105
9132257787327384
1031150500504133
119a1a94249627a
1235b1213711b71
13144c63088791a
1479999982bb13
1539046903723d
hex1c54ca846245

31150500504133 has 2 divisors, whose sum is σ = 31150500504134. Its totient is φ = 31150500504132.

The previous prime is 31150500504041. The next prime is 31150500504161. The reversal of 31150500504133 is 33140500505113.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 25763004578529 + 5387495925604 = 5075727^2 + 2321098^2 .

It is an emirp because it is prime and its reverse (33140500505113) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 31150500504133 - 29 = 31150500503621 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 31150500504095 and 31150500504104.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (31150500504173) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 15575250252066 + 15575250252067.

It is an arithmetic number, because the mean of its divisors is an integer number (15575250252067).

Almost surely, 231150500504133 is an apocalyptic number.

It is an amenable number.

31150500504133 is a deficient number, since it is larger than the sum of its proper divisors (1).

31150500504133 is an equidigital number, since it uses as much as digits as its factorization.

31150500504133 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 13500, while the sum is 31.

The spelling of 31150500504133 in words is "thirty-one trillion, one hundred fifty billion, five hundred million, five hundred four thousand, one hundred thirty-three".