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132550623433 is a prime number
BaseRepresentation
bin111101101110010100…
…0100000000011001001
3110200010200101100012101
41323130220200003021
54132430424422213
6140520511513401
712401504465164
oct1733450400311
9420120340171
10132550623433
115123a40a288
1221832b28861
13c66544a077
1465b6136adb
1536abc3dbdd
hex1edca200c9

132550623433 has 2 divisors, whose sum is σ = 132550623434. Its totient is φ = 132550623432.

The previous prime is 132550623431. The next prime is 132550623461. The reversal of 132550623433 is 334326055231.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 132519297024 + 31326409 = 364032^2 + 5597^2 .

It is a cyclic number.

It is not a de Polignac number, because 132550623433 - 21 = 132550623431 is a prime.

Together with 132550623431, it forms a pair of twin primes.

It is a junction number, because it is equal to n+sod(n) for n = 132550623392 and 132550623401.

It is not a weakly prime, because it can be changed into another prime (132550623431) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 66275311716 + 66275311717.

It is an arithmetic number, because the mean of its divisors is an integer number (66275311717).

Almost surely, 2132550623433 is an apocalyptic number.

It is an amenable number.

132550623433 is a deficient number, since it is larger than the sum of its proper divisors (1).

132550623433 is an equidigital number, since it uses as much as digits as its factorization.

132550623433 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 194400, while the sum is 37.

Adding to 132550623433 its reverse (334326055231), we get a palindrome (466876678664).

The spelling of 132550623433 in words is "one hundred thirty-two billion, five hundred fifty million, six hundred twenty-three thousand, four hundred thirty-three".