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1100001122153 is a prime number
BaseRepresentation
bin10000000000011101001…
…011010001011101101001
310220011021222011210000222
4100000131023101131221
5121010300241402103
62201200013335425
7142321024160063
oct20003513213551
93804258153028
101100001122153
11394564a5809a
12159230197575
137c9640abc7b
143b3516ba733
151d930b23638
hex1001d2d1769

1100001122153 has 2 divisors, whose sum is σ = 1100001122154. Its totient is φ = 1100001122152.

The previous prime is 1100001122117. The next prime is 1100001122159. The reversal of 1100001122153 is 3512211000011.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 718920843664 + 381080278489 = 847892^2 + 617317^2 .

It is a cyclic number.

It is not a de Polignac number, because 1100001122153 - 210 = 1100001121129 is a prime.

It is not a weakly prime, because it can be changed into another prime (1100001122159) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (17) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 550000561076 + 550000561077.

It is an arithmetic number, because the mean of its divisors is an integer number (550000561077).

Almost surely, 21100001122153 is an apocalyptic number.

It is an amenable number.

1100001122153 is a deficient number, since it is larger than the sum of its proper divisors (1).

1100001122153 is an equidigital number, since it uses as much as digits as its factorization.

1100001122153 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 60, while the sum is 17.

Adding to 1100001122153 its reverse (3512211000011), we get a palindrome (4612212122164).

The spelling of 1100001122153 in words is "one trillion, one hundred billion, one million, one hundred twenty-two thousand, one hundred fifty-three".