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131403312132401 is a prime number
BaseRepresentation
bin11101111000001010110111…
…101111011111000100110001
3122020021000220202100211202002
4131320022313233133010301
5114210402440401214101
61143245503115103345
736451401400054301
oct3570126757370461
9566230822324662
10131403312132401
1138961917026621
12128a2a13441b55
13584238242b6ca
142463d50621601
15102d17d3a5d6b
hex7782b7bdf131

131403312132401 has 2 divisors, whose sum is σ = 131403312132402. Its totient is φ = 131403312132400.

The previous prime is 131403312132361. The next prime is 131403312132419. The reversal of 131403312132401 is 104231213304131.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 102560665946176 + 28842646186225 = 10127224^2 + 5370535^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-131403312132401 is a prime.

It is not a weakly prime, because it can be changed into another prime (131403312162401) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65701656066200 + 65701656066201.

It is an arithmetic number, because the mean of its divisors is an integer number (65701656066201).

Almost surely, 2131403312132401 is an apocalyptic number.

It is an amenable number.

131403312132401 is a deficient number, since it is larger than the sum of its proper divisors (1).

131403312132401 is an equidigital number, since it uses as much as digits as its factorization.

131403312132401 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 5184, while the sum is 29.

Adding to 131403312132401 its reverse (104231213304131), we get a palindrome (235634525436532).

The spelling of 131403312132401 in words is "one hundred thirty-one trillion, four hundred three billion, three hundred twelve million, one hundred thirty-two thousand, four hundred one".