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201101112000251 is a prime number
BaseRepresentation
bin101101101110011010000000…
…010110000011111011111011
3222101001002010021010021212122
4231232122000112003323323
5202324320034133002001
61551412324145031455
760234035012015351
oct5556320026037373
9871032107107778
10201101112000251
1159093559035611
121a67a8ab62558b
1388299a4559c68
1437934baa717d1
15183b17c496b1b
hexb6e680583efb

201101112000251 has 2 divisors, whose sum is σ = 201101112000252. Its totient is φ = 201101112000250.

The previous prime is 201101112000199. The next prime is 201101112000259. The reversal of 201101112000251 is 152000211101102.

It is a strong prime.

It is a cyclic number.

It is not a de Polignac number, because 201101112000251 - 222 = 201101107805947 is a prime.

It is a super-3 number, since 3×2011011120002513 (a number of 44 digits) contains 333 as substring.

It is a Sophie Germain prime.

It is not a weakly prime, because it can be changed into another prime (201101112000259) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 100550556000125 + 100550556000126.

It is an arithmetic number, because the mean of its divisors is an integer number (100550556000126).

Almost surely, 2201101112000251 is an apocalyptic number.

201101112000251 is a deficient number, since it is larger than the sum of its proper divisors (1).

201101112000251 is an equidigital number, since it uses as much as digits as its factorization.

201101112000251 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 40, while the sum is 17.

Adding to 201101112000251 its reverse (152000211101102), we get a palindrome (353101323101353).

The spelling of 201101112000251 in words is "two hundred one trillion, one hundred one billion, one hundred twelve million, two hundred fifty-one", and thus it is an aban number.