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1100122112147 is a prime number
BaseRepresentation
bin10000000000100100011…
…000110100000010010011
310220011121101211201002212
4100000210120310002103
5121011022230042042
62201220014502335
7142324023443432
oct20004430640223
93804541751085
101100122112147
11394617285723
121592648049ab
137c983191696
143b3637b3119
151d93b572482
hex10024634093

1100122112147 has 2 divisors, whose sum is σ = 1100122112148. Its totient is φ = 1100122112146.

The previous prime is 1100122112107. The next prime is 1100122112149. The reversal of 1100122112147 is 7412112210011.

It is a strong prime.

It is a cyclic number.

It is not a de Polignac number, because 1100122112147 - 26 = 1100122112083 is a prime.

Together with 1100122112149, it forms a pair of twin primes.

It is a Chen prime.

It is not a weakly prime, because it can be changed into another prime (1100122112149) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 550061056073 + 550061056074.

It is an arithmetic number, because the mean of its divisors is an integer number (550061056074).

Almost surely, 21100122112147 is an apocalyptic number.

1100122112147 is a deficient number, since it is larger than the sum of its proper divisors (1).

1100122112147 is an equidigital number, since it uses as much as digits as its factorization.

1100122112147 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 224, while the sum is 23.

Adding to 1100122112147 its reverse (7412112210011), we get a palindrome (8512234322158).

The spelling of 1100122112147 in words is "one trillion, one hundred billion, one hundred twenty-two million, one hundred twelve thousand, one hundred forty-seven".