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100011132437 is a prime number
BaseRepresentation
bin101110100100100100…
…0001100011000010101
3100120010221120221002012
41131021020030120111
53114310322214222
6113540001354005
710140240412406
oct1351110143025
9316127527065
10100011132437
113946174a067
1217471638905
13957ac09898
144baa71b9ad
1529051d81e2
hex174920c615

100011132437 has 2 divisors, whose sum is σ = 100011132438. Its totient is φ = 100011132436.

The previous prime is 100011132293. The next prime is 100011132451. The reversal of 100011132437 is 734231110001.

It is a happy number.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 65399878756 + 34611253681 = 255734^2 + 186041^2 .

It is a cyclic number.

It is not a de Polignac number, because 100011132437 - 210 = 100011131413 is a prime.

It is a super-2 number, since 2×1000111324372 (a number of 23 digits) contains 22 as substring.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (100011132457) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 50005566218 + 50005566219.

It is an arithmetic number, because the mean of its divisors is an integer number (50005566219).

Almost surely, 2100011132437 is an apocalyptic number.

It is an amenable number.

100011132437 is a deficient number, since it is larger than the sum of its proper divisors (1).

100011132437 is an equidigital number, since it uses as much as digits as its factorization.

100011132437 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 504, while the sum is 23.

Adding to 100011132437 its reverse (734231110001), we get a palindrome (834242242438).

The spelling of 100011132437 in words is "one hundred billion, eleven million, one hundred thirty-two thousand, four hundred thirty-seven".