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21000210000113 is a prime number
BaseRepresentation
bin1001100011001011111100…
…00011101010100011110001
32202100121012202001001010202
410301211332003222203301
510223031412230000423
6112355210245404545
74265133531525101
oct461457603524361
982317182031122
1021000210000113
116767159086143
122431b93a09755
13b94407c284a6
145285b652b201
152663e4c7e728
hex13197e0ea8f1

21000210000113 has 2 divisors, whose sum is σ = 21000210000114. Its totient is φ = 21000210000112.

The previous prime is 21000210000101. The next prime is 21000210000173. The reversal of 21000210000113 is 31100001200012.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 17658417604864 + 3341792395249 = 4202192^2 + 1828057^2 .

It is a cyclic number.

It is not a de Polignac number, because 21000210000113 - 24 = 21000210000097 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 21000210000094 and 21000210000103.

It is not a weakly prime, because it can be changed into another prime (21000210000173) 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, 10500105000056 + 10500105000057.

It is an arithmetic number, because the mean of its divisors is an integer number (10500105000057).

Almost surely, 221000210000113 is an apocalyptic number.

It is an amenable number.

21000210000113 is a deficient number, since it is larger than the sum of its proper divisors (1).

21000210000113 is an equidigital number, since it uses as much as digits as its factorization.

21000210000113 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 12, while the sum is 11.

Adding to 21000210000113 its reverse (31100001200012), we get a palindrome (52100211200125).

The spelling of 21000210000113 in words is "twenty-one trillion, two hundred ten million, one hundred thirteen", and thus it is an aban number.