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1301003312173 is a prime number
BaseRepresentation
bin10010111011101001110…
…101110001110000101101
311121101010011020222210021
4102323221311301300231
5132303423321442143
62433401144042141
7162665034455635
oct22735165616055
94541104228707
101301003312173
114618304a8504
12190187463351
13958b7c24096
1446d7c9196c5
1523c9705cced
hex12ee9d71c2d

1301003312173 has 2 divisors, whose sum is σ = 1301003312174. Its totient is φ = 1301003312172.

The previous prime is 1301003312057. The next prime is 1301003312221. The reversal of 1301003312173 is 3712133001031.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 1294891512489 + 6111799684 = 1137933^2 + 78178^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-1301003312173 is a prime.

It is a super-3 number, since 3×13010033121733 (a number of 37 digits) contains 333 as substring.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (1301003322173) 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, 650501656086 + 650501656087.

It is an arithmetic number, because the mean of its divisors is an integer number (650501656087).

Almost surely, 21301003312173 is an apocalyptic number.

It is an amenable number.

1301003312173 is a deficient number, since it is larger than the sum of its proper divisors (1).

1301003312173 is an equidigital number, since it uses as much as digits as its factorization.

1301003312173 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 1134, while the sum is 25.

The spelling of 1301003312173 in words is "one trillion, three hundred one billion, three million, three hundred twelve thousand, one hundred seventy-three".