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1355416550473 is a prime number
BaseRepresentation
bin10011101110010101000…
…111110011110001001001
311210120120022100220020111
4103232111013303301021
5134201343114103343
62514400400354321
7166632325404436
oct23562507636111
94716508326214
101355416550473
11482913341837
1219a8323159a1
139aa7a0b05cb
144986140cb8d
15253ce0a3c9d
hex13b951f3c49

1355416550473 has 2 divisors, whose sum is σ = 1355416550474. Its totient is φ = 1355416550472.

The previous prime is 1355416550471. The next prime is 1355416550563. The reversal of 1355416550473 is 3740556145531.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 1317690480649 + 37726069824 = 1147907^2 + 194232^2 .

It is a cyclic number.

It is not a de Polignac number, because 1355416550473 - 21 = 1355416550471 is a prime.

It is a super-2 number, since 2×13554165504732 (a number of 25 digits) contains 22 as substring.

Together with 1355416550471, it forms a pair of twin primes.

It is not a weakly prime, because it can be changed into another prime (1355416550471) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 677708275236 + 677708275237.

It is an arithmetic number, because the mean of its divisors is an integer number (677708275237).

Almost surely, 21355416550473 is an apocalyptic number.

It is an amenable number.

1355416550473 is a deficient number, since it is larger than the sum of its proper divisors (1).

1355416550473 is an equidigital number, since it uses as much as digits as its factorization.

1355416550473 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 3780000, while the sum is 49.

The spelling of 1355416550473 in words is "one trillion, three hundred fifty-five billion, four hundred sixteen million, five hundred fifty thousand, four hundred seventy-three".