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136000536254657 is a prime number
BaseRepresentation
bin11110111011000100010111…
…101001110001100011000001
3122211112112011012121101202122
4132323010113221301203001
5120311213044240122112
61201125441150454025
740434500033255414
oct3673042751614301
9584475135541678
10136000536254657
113a37455a209748
1213305994504315
135ab6a45a51c19
14258266529db7b
1510aca4565b872
hex7bb117a718c1

136000536254657 has 2 divisors, whose sum is σ = 136000536254658. Its totient is φ = 136000536254656.

The previous prime is 136000536254593. The next prime is 136000536254711. The reversal of 136000536254657 is 756452635000631.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 110680478389441 + 25320057865216 = 10520479^2 + 5031904^2 .

It is a cyclic number.

It is not a de Polignac number, because 136000536254657 - 26 = 136000536254593 is a prime.

It is a super-2 number, since 2×1360005362546572 (a number of 29 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 136000536254599 and 136000536254608.

It is not a weakly prime, because it can be changed into another prime (136000536254857) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 68000268127328 + 68000268127329.

It is an arithmetic number, because the mean of its divisors is an integer number (68000268127329).

Almost surely, 2136000536254657 is an apocalyptic number.

It is an amenable number.

136000536254657 is a deficient number, since it is larger than the sum of its proper divisors (1).

136000536254657 is an equidigital number, since it uses as much as digits as its factorization.

136000536254657 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 13608000, while the sum is 53.

The spelling of 136000536254657 in words is "one hundred thirty-six trillion, five hundred thirty-six million, two hundred fifty-four thousand, six hundred fifty-seven".