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131925496433 is a prime number
BaseRepresentation
bin111101011011101011…
…1110101001001110001
3110121112010002121011112
41322313113311021301
54130140411341213
6140334501123105
712350144134403
oct1726727651161
9417463077145
10131925496433
1150a4955a0a5
12216996b9495
13c595a94533
1465570ccb73
153671e0aea8
hex1eb75f5271

131925496433 has 2 divisors, whose sum is σ = 131925496434. Its totient is φ = 131925496432.

The previous prime is 131925496429. The next prime is 131925496441. The reversal of 131925496433 is 334694529131.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 104295410704 + 27630085729 = 322948^2 + 166223^2 .

It is an emirp because it is prime and its reverse (334694529131) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 131925496433 - 22 = 131925496429 is a prime.

It is not a weakly prime, because it can be changed into another prime (131925496033) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (23) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65962748216 + 65962748217.

It is an arithmetic number, because the mean of its divisors is an integer number (65962748217).

Almost surely, 2131925496433 is an apocalyptic number.

It is an amenable number.

131925496433 is a deficient number, since it is larger than the sum of its proper divisors (1).

131925496433 is an equidigital number, since it uses as much as digits as its factorization.

131925496433 is an odious number, because the sum of its binary digits is odd.

The product of its digits is 2099520, while the sum is 50.

The spelling of 131925496433 in words is "one hundred thirty-one billion, nine hundred twenty-five million, four hundred ninety-six thousand, four hundred thirty-three".