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100804148071433 is a prime number
BaseRepresentation
bin10110111010111001001011…
…011100000000110000001001
3111012220210010101210010112122
4112322321023130000300021
5101203033343401241213
6554220441502421025
730142600522026062
oct2672711334006011
9435823111703478
10100804148071433
112a13488769a384
12b380629291775
134432a375b9151
141ac6d38033569
15b9c231d34d08
hex5bae4b700c09

100804148071433 has 2 divisors, whose sum is σ = 100804148071434. Its totient is φ = 100804148071432.

The previous prime is 100804148071339. The next prime is 100804148071447. The reversal of 100804148071433 is 334170841408001.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 52095655223824 + 48708492847609 = 7217732^2 + 6979147^2 .

It is a cyclic number.

It is not a de Polignac number, because 100804148071433 - 28 = 100804148071177 is a prime.

It is a super-2 number, since 2×1008041480714332 (a number of 29 digits) contains 22 as substring.

It is not a weakly prime, because it can be changed into another prime (100804148071633) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 50402074035716 + 50402074035717.

It is an arithmetic number, because the mean of its divisors is an integer number (50402074035717).

Almost surely, 2100804148071433 is an apocalyptic number.

It is an amenable number.

100804148071433 is a deficient number, since it is larger than the sum of its proper divisors (1).

100804148071433 is an equidigital number, since it uses as much as digits as its factorization.

100804148071433 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 258048, while the sum is 44.

Adding to 100804148071433 its reverse (334170841408001), we get a palindrome (434974989479434).

The spelling of 100804148071433 in words is "one hundred trillion, eight hundred four billion, one hundred forty-eight million, seventy-one thousand, four hundred thirty-three".