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995981340201703 is a prime number
BaseRepresentation
bin1110001001110101101111100…
…1111111000101001011100111
311211121110212221001202121210021
43202131123321333011023213
52021021124241042423303
613450135302225140011
7416534164041431152
oct34235337177051347
94747425831677707
10995981340201703
11269394350400369
1293857a3378a607
133399883942688b
14137d3a7b278c99
157a22b610a6ebd
hex389d6f9fc52e7

995981340201703 has 2 divisors, whose sum is σ = 995981340201704. Its totient is φ = 995981340201702.

The previous prime is 995981340201557. The next prime is 995981340201721. The reversal of 995981340201703 is 307102043189599.

It is a strong prime.

It is a cyclic number.

It is not a de Polignac number, because 995981340201703 - 217 = 995981340070631 is a prime.

It is a super-3 number, since 3×9959813402017033 (a number of 46 digits) contains 333 as substring.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (995981340201763) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (31) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 497990670100851 + 497990670100852.

It is an arithmetic number, because the mean of its divisors is an integer number (497990670100852).

Almost surely, 2995981340201703 is an apocalyptic number.

995981340201703 is a deficient number, since it is larger than the sum of its proper divisors (1).

995981340201703 is an equidigital number, since it uses as much as digits as its factorization.

995981340201703 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 14696640, while the sum is 61.

The spelling of 995981340201703 in words is "nine hundred ninety-five trillion, nine hundred eighty-one billion, three hundred forty million, two hundred one thousand, seven hundred three".