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551491233941 is a prime number
BaseRepresentation
bin10000000011001110111…
…00000110100010010101
31221201111101001102021002
420001213130012202111
533013423223441231
61101203535033045
754562316100044
oct10014734064225
91851441042232
10551491233941
111a298247369a
128aa7136b785
134000bb30a3b
141c999a1455b
15e52b3900cb
hex8067706895

551491233941 has 2 divisors, whose sum is σ = 551491233942. Its totient is φ = 551491233940.

The previous prime is 551491233863. The next prime is 551491233959. The reversal of 551491233941 is 149332194155.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 483511622500 + 67979611441 = 695350^2 + 260729^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-551491233941 is a prime.

It is a super-3 number, since 3×5514912339413 (a number of 36 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (551491233971) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 275745616970 + 275745616971.

It is an arithmetic number, because the mean of its divisors is an integer number (275745616971).

Almost surely, 2551491233941 is an apocalyptic number.

It is an amenable number.

551491233941 is a deficient number, since it is larger than the sum of its proper divisors (1).

551491233941 is an equidigital number, since it uses as much as digits as its factorization.

551491233941 is an evil number, because the sum of its binary digits is even.

The product of its digits is 583200, while the sum is 47.

The spelling of 551491233941 in words is "five hundred fifty-one billion, four hundred ninety-one million, two hundred thirty-three thousand, nine hundred forty-one".