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121145302011077 is a prime number
BaseRepresentation
bin11011100010111001010110…
…101100010100110011000101
3120212221101012101120102001022
4123202321112230110303011
5111334321034303323302
61105353214453440525
734342313342513561
oct3342712654246305
9525841171512038
10121145302011077
1135667485808683
1211706930640145
135279c50869356
1421cb66cd406a1
15e014022ee8a2
hex6e2e56b14cc5

121145302011077 has 2 divisors, whose sum is σ = 121145302011078. Its totient is φ = 121145302011076.

The previous prime is 121145302011037. The next prime is 121145302011121. The reversal of 121145302011077 is 770110203541121.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 73745946302116 + 47399355708961 = 8587546^2 + 6884719^2 .

It is a cyclic number.

It is not a de Polignac number, because 121145302011077 - 210 = 121145302010053 is a prime.

It is a super-3 number, since 3×1211453020110773 (a number of 43 digits) contains 333 as substring.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (121145302011037) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 60572651005538 + 60572651005539.

It is an arithmetic number, because the mean of its divisors is an integer number (60572651005539).

Almost surely, 2121145302011077 is an apocalyptic number.

It is an amenable number.

121145302011077 is a deficient number, since it is larger than the sum of its proper divisors (1).

121145302011077 is an equidigital number, since it uses as much as digits as its factorization.

121145302011077 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 11760, while the sum is 35.

Adding to 121145302011077 its reverse (770110203541121), we get a palindrome (891255505552198).

The spelling of 121145302011077 in words is "one hundred twenty-one trillion, one hundred forty-five billion, three hundred two million, eleven thousand, seventy-seven".