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135445212441193 is a prime number
BaseRepresentation
bin11110110010111111001011…
…110001100010101001101001
3122202120102202201212202202211
4132302333023301202221221
5120223113243341104233
61200022401022125121
740346410425605164
oct3662771361425151
9582512681782684
10135445212441193
113a17aaa0412366
12132362337207a1
135a76575bba965
142563824676adb
15109d392ab75cd
hex7b2fcbc62a69

135445212441193 has 2 divisors, whose sum is σ = 135445212441194. Its totient is φ = 135445212441192.

The previous prime is 135445212441169. The next prime is 135445212441263. The reversal of 135445212441193 is 391144212544531.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 107871654832384 + 27573557608809 = 10386128^2 + 5251053^2 .

It is an emirp because it is prime and its reverse (391144212544531) is a distict prime.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-135445212441193 is a prime.

It is not a weakly prime, because it can be changed into another prime (135445212441593) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 67722606220596 + 67722606220597.

It is an arithmetic number, because the mean of its divisors is an integer number (67722606220597).

Almost surely, 2135445212441193 is an apocalyptic number.

It is an amenable number.

135445212441193 is a deficient number, since it is larger than the sum of its proper divisors (1).

135445212441193 is an equidigital number, since it uses as much as digits as its factorization.

135445212441193 is an odious number, because the sum of its binary digits is odd.

The product of its digits is 2073600, while the sum is 49.

The spelling of 135445212441193 in words is "one hundred thirty-five trillion, four hundred forty-five billion, two hundred twelve million, four hundred forty-one thousand, one hundred ninety-three".