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102022112012153 is a prime number
BaseRepresentation
bin10111001100100111011111…
…101111101110011101111001
3111101020012220101210222111212
4113030213133233232131321
5101333012241133342103
61000552151120532505
730326566020511124
oct2714473757563571
9441205811728455
10102022112012153
112a564375579a73
12b53869aa81135
1344c084a7ba478
141b29c7a27d3bb
15bbdc68b547d8
hex5cc9dfbee779

102022112012153 has 2 divisors, whose sum is σ = 102022112012154. Its totient is φ = 102022112012152.

The previous prime is 102022112012071. The next prime is 102022112012189. The reversal of 102022112012153 is 351210211220201.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 97539563468944 + 4482548543209 = 9876212^2 + 2117203^2 .

It is a cyclic number.

It is not a de Polignac number, because 102022112012153 - 222 = 102022107817849 is a prime.

It is a super-3 number, since 3×1020221120121533 (a number of 43 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.

It is not a weakly prime, because it can be changed into another prime (102022112011153) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 51011056006076 + 51011056006077.

It is an arithmetic number, because the mean of its divisors is an integer number (51011056006077).

Almost surely, 2102022112012153 is an apocalyptic number.

It is an amenable number.

102022112012153 is a deficient number, since it is larger than the sum of its proper divisors (1).

102022112012153 is an equidigital number, since it uses as much as digits as its factorization.

102022112012153 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 480, while the sum is 23.

Adding to 102022112012153 its reverse (351210211220201), we get a palindrome (453232323232354).

The spelling of 102022112012153 in words is "one hundred two trillion, twenty-two billion, one hundred twelve million, twelve thousand, one hundred fifty-three".