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332124144020131 is a prime number
BaseRepresentation
bin100101110000100001010110…
…1101000000010101010100011
31121112221202110122100211201211
41023201002231220002222203
5322013010221332121011
63134211420405410551
7126646114621346161
oct11341025550025243
91545852418324654
10332124144020131
1196909070928101
12312bba73a09a57
131134223199b2b8
145c02c740dc231
15285e4a02ddb21
hex12e10ada02aa3

332124144020131 has 2 divisors, whose sum is σ = 332124144020132. Its totient is φ = 332124144020130.

The previous prime is 332124144020111. The next prime is 332124144020219. The reversal of 332124144020131 is 131020441421233.

It is a happy number.

It is a weak prime.

It is a cyclic number.

It is not a de Polignac number, because 332124144020131 - 211 = 332124144018083 is a prime.

It is a super-2 number, since 2×3321241440201312 (a number of 30 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 332124144020093 and 332124144020102.

It is not a weakly prime, because it can be changed into another prime (332124144020111) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 166062072010065 + 166062072010066.

It is an arithmetic number, because the mean of its divisors is an integer number (166062072010066).

Almost surely, 2332124144020131 is an apocalyptic number.

332124144020131 is a deficient number, since it is larger than the sum of its proper divisors (1).

332124144020131 is an equidigital number, since it uses as much as digits as its factorization.

332124144020131 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 13824, while the sum is 31.

Adding to 332124144020131 its reverse (131020441421233), we get a palindrome (463144585441364).

The spelling of 332124144020131 in words is "three hundred thirty-two trillion, one hundred twenty-four billion, one hundred forty-four million, twenty thousand, one hundred thirty-one".