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13121001130001 is a prime number
BaseRepresentation
bin1011111011101111100010…
…0111001000100000010001
31201110100121010110210221212
42332323320213020200101
53204433302242130001
643523412114425505
72522650432262222
oct276737047104021
951410533423855
1013121001130001
1141a964908a686
12157ab30892295
137423c73b70ab
143350ba2d1249
1517b4924937bb
hexbeef89c8811

13121001130001 has 2 divisors, whose sum is σ = 13121001130002. Its totient is φ = 13121001130000.

The previous prime is 13121001129991. The next prime is 13121001130033. The reversal of 13121001130001 is 10003110012131.

It is a happy number.

13121001130001 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 9775321155601 + 3345679974400 = 3126551^2 + 1829120^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-13121001130001 is a prime.

It is not a weakly prime, because it can be changed into another prime (13121001130901) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6560500565000 + 6560500565001.

It is an arithmetic number, because the mean of its divisors is an integer number (6560500565001).

Almost surely, 213121001130001 is an apocalyptic number.

It is an amenable number.

13121001130001 is a deficient number, since it is larger than the sum of its proper divisors (1).

13121001130001 is an equidigital number, since it uses as much as digits as its factorization.

13121001130001 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 18, while the sum is 14.

Adding to 13121001130001 its reverse (10003110012131), we get a palindrome (23124111142132).

The spelling of 13121001130001 in words is "thirteen trillion, one hundred twenty-one billion, one million, one hundred thirty thousand, one".