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132052510311151 = 393233358149437
BaseRepresentation
bin11110000001100111011110…
…111110000000001011101111
3122022120002121111021022212211
4132001213132332000023233
5114302022020114424101
61144504030305111251
736546323210404633
oct3601473676001357
9568502544238784
10132052510311151
1139092178517a81
12129887b2628527
13588b662988a91
142487538914dc3
15103eec74d1c51
hex7819def802ef

132052510311151 has 4 divisors (see below), whose sum is σ = 132055868499912. Its totient is φ = 132049152122392.

The previous prime is 132052510311053. The next prime is 132052510311181. The reversal of 132052510311151 is 151113015250231.

It is a semiprime because it is the product of two primes.

It is a cyclic number.

It is not a de Polignac number, because 132052510311151 - 221 = 132052508213999 is a prime.

It is a Duffinian number.

It is a congruent number.

It is not an unprimeable number, because it can be changed into a prime (132052510311181) by changing a digit.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 1679035396 + ... + 1679114041.

It is an arithmetic number, because the mean of its divisors is an integer number (33013967124978).

Almost surely, 2132052510311151 is an apocalyptic number.

132052510311151 is a deficient number, since it is larger than the sum of its proper divisors (3358188761).

132052510311151 is an equidigital number, since it uses as much as digits as its factorization.

132052510311151 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 3358188760.

The product of its (nonzero) digits is 4500, while the sum is 31.

Adding to 132052510311151 its reverse (151113015250231), we get a palindrome (283165525561382).

The spelling of 132052510311151 in words is "one hundred thirty-two trillion, fifty-two billion, five hundred ten million, three hundred eleven thousand, one hundred fifty-one".

Divisors: 1 39323 3358149437 132052510311151