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530110300213997 is a prime number
BaseRepresentation
bin111100010001000011110110…
…0001000101101001011101101
32120111222001100021120211110122
41320202013230020231023231
51023440311343323321442
65115241150541214325
7216442130264035046
oct17042075410551355
92514861307524418
10530110300213997
11143a00574584178
124b556b16b613a5
13199a4257574149
1494c96928178cd
1541445bad21dd2
hex1e221ec22d2ed

530110300213997 has 2 divisors, whose sum is σ = 530110300213998. Its totient is φ = 530110300213996.

The previous prime is 530110300213961. The next prime is 530110300214059. The reversal of 530110300213997 is 799312003011035.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 300780134604196 + 229330165609801 = 17343014^2 + 15143651^2 .

It is a cyclic number.

It is not a de Polignac number, because 530110300213997 - 212 = 530110300209901 is a prime.

It is a super-3 number, since 3×5301103002139973 (a number of 45 digits) contains 333 as substring.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (530110300213397) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 265055150106998 + 265055150106999.

It is an arithmetic number, because the mean of its divisors is an integer number (265055150106999).

Almost surely, 2530110300213997 is an apocalyptic number.

It is an amenable number.

530110300213997 is a deficient number, since it is larger than the sum of its proper divisors (1).

530110300213997 is an equidigital number, since it uses as much as digits as its factorization.

530110300213997 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 153090, while the sum is 44.

The spelling of 530110300213997 in words is "five hundred thirty trillion, one hundred ten billion, three hundred million, two hundred thirteen thousand, nine hundred ninety-seven".