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13124763279401 is a prime number
BaseRepresentation
bin1011111011111101100011…
…0110100110000000101001
31201110201022022121210201112
42332333120312212000221
53210014003344420101
643525233310400105
72523142604104634
oct276773066460051
951421268553645
1013124763279401
1142001a9789164
12157b800800035
1374287695b0b7
14335355c0cc1b
1517b6128d46bb
hexbefd8da6029

13124763279401 has 2 divisors, whose sum is σ = 13124763279402. Its totient is φ = 13124763279400.

The previous prime is 13124763279337. The next prime is 13124763279487. The reversal of 13124763279401 is 10497236742131.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 11492371201600 + 1632392077801 = 3390040^2 + 1277651^2 .

It is a cyclic number.

It is not a de Polignac number, because 13124763279401 - 26 = 13124763279337 is a prime.

It is a super-2 number, since 2×131247632794012 (a number of 27 digits) contains 22 as substring.

It is a self number, because there is not a number n which added to its sum of digits gives 13124763279401.

It is not a weakly prime, because it can be changed into another prime (13124763279101) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6562381639700 + 6562381639701.

It is an arithmetic number, because the mean of its divisors is an integer number (6562381639701).

Almost surely, 213124763279401 is an apocalyptic number.

It is an amenable number.

13124763279401 is a deficient number, since it is larger than the sum of its proper divisors (1).

13124763279401 is an equidigital number, since it uses as much as digits as its factorization.

13124763279401 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 1524096, while the sum is 50.

The spelling of 13124763279401 in words is "thirteen trillion, one hundred twenty-four billion, seven hundred sixty-three million, two hundred seventy-nine thousand, four hundred one".