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141202012103011 is a prime number
BaseRepresentation
bin100000000110110000100111…
…110001011110000101100011
3200111221210002000002002000101
4200012300213301132011203
5122001423210044244021
61220151150414003231
741512336330422466
oct4006604761360543
9614853060062011
10141202012103011
1140a9a489781092
1213a05a901a6517
1360a33a5b1c5b2
1426c22dcb412dd
15114cec5b7aa91
hex806c27c5e163

141202012103011 has 2 divisors, whose sum is σ = 141202012103012. Its totient is φ = 141202012103010.

The previous prime is 141202012102951. The next prime is 141202012103021. The reversal of 141202012103011 is 110301210202141.

It is a strong prime.

It is a cyclic number.

It is not a de Polignac number, because 141202012103011 - 29 = 141202012102499 is a prime.

It is a super-3 number, since 3×1412020121030113 (a number of 43 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.

It is a self number, because there is not a number n which added to its sum of digits gives 141202012103011.

It is not a weakly prime, because it can be changed into another prime (141202012103021) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 70601006051505 + 70601006051506.

It is an arithmetic number, because the mean of its divisors is an integer number (70601006051506).

Almost surely, 2141202012103011 is an apocalyptic number.

141202012103011 is a deficient number, since it is larger than the sum of its proper divisors (1).

141202012103011 is an equidigital number, since it uses as much as digits as its factorization.

141202012103011 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 96, while the sum is 19.

Adding to 141202012103011 its reverse (110301210202141), we get a palindrome (251503222305152).

The spelling of 141202012103011 in words is "one hundred forty-one trillion, two hundred two billion, twelve million, one hundred three thousand, eleven".