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11212111031293 is a prime number
BaseRepresentation
bin1010001100101000010111…
…1001010010101111111101
31110200212102021222111110101
42203022011321102233331
52432144400411000133
635502434523041101
72235022364202424
oct243120571225775
943625367874411
1011212111031293
113633036418187
121310b95370191
136343b38b1623
142aa951cd82bb
151469bced237d
hexa3285e52bfd

11212111031293 has 2 divisors, whose sum is σ = 11212111031294. Its totient is φ = 11212111031292.

The previous prime is 11212111031263. The next prime is 11212111031321. The reversal of 11212111031293 is 39213011121211.

It is an a-pointer prime, because the next prime (11212111031321) can be obtained adding 11212111031293 to its sum of digits (28).

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 10468829100249 + 743281931044 = 3235557^2 + 862138^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-11212111031293 is a prime.

It is a super-2 number, since 2×112121110312932 (a number of 27 digits) contains 22 as substring.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (11212111031263) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5606055515646 + 5606055515647.

It is an arithmetic number, because the mean of its divisors is an integer number (5606055515647).

Almost surely, 211212111031293 is an apocalyptic number.

It is an amenable number.

11212111031293 is a deficient number, since it is larger than the sum of its proper divisors (1).

11212111031293 is an equidigital number, since it uses as much as digits as its factorization.

11212111031293 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 648, while the sum is 28.

The spelling of 11212111031293 in words is "eleven trillion, two hundred twelve billion, one hundred eleven million, thirty-one thousand, two hundred ninety-three".