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683116551121 is a prime number
BaseRepresentation
bin10011111000011001110…
…10110101101111010001
32102022020120022022220111
421330030322311233101
542143010314113441
61241452554241321
7100232156610334
oct11741472655721
92368216268814
10683116551121
11243787536664
12b0486426241
134c5576639c4
14250c4cc461b
1512b81c54081
hex9f0ceb5bd1

683116551121 has 2 divisors, whose sum is σ = 683116551122. Its totient is φ = 683116551120.

The previous prime is 683116551083. The next prime is 683116551127. The reversal of 683116551121 is 121155611386.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 642304470721 + 40812080400 = 801439^2 + 202020^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-683116551121 is a prime.

It is a super-4 number, since 4×6831165511214 (a number of 48 digits) contains 4444 as substring.

It is not a weakly prime, because it can be changed into another prime (683116551127) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (23) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 341558275560 + 341558275561.

It is an arithmetic number, because the mean of its divisors is an integer number (341558275561).

Almost surely, 2683116551121 is an apocalyptic number.

It is an amenable number.

683116551121 is a deficient number, since it is larger than the sum of its proper divisors (1).

683116551121 is an equidigital number, since it uses as much as digits as its factorization.

683116551121 is an odious number, because the sum of its binary digits is odd.

The product of its digits is 43200, while the sum is 40.

The spelling of 683116551121 in words is "six hundred eighty-three billion, one hundred sixteen million, five hundred fifty-one thousand, one hundred twenty-one".