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131001341114093 is a prime number
BaseRepresentation
bin11101110010010100100000…
…011001101111011011101101
3122011211121100200222100221012
4131302110200121233123231
5114132311221311122333
61142341104005115005
736410351232556232
oct3562244031573355
9564747320870835
10131001341114093
11388173a1951a7a
1212838b300b8a65
1358134b1652658
14244c6da789789
1510229a3921e48
hex77252066f6ed

131001341114093 has 2 divisors, whose sum is σ = 131001341114094. Its totient is φ = 131001341114092.

The previous prime is 131001341114029. The next prime is 131001341114113. The reversal of 131001341114093 is 390411143100131.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 130930051046089 + 71290068004 = 11442467^2 + 267002^2 .

It is an emirp because it is prime and its reverse (390411143100131) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 131001341114093 - 26 = 131001341114029 is a prime.

It is a super-2 number, since 2×1310013411140932 (a number of 29 digits) contains 22 as substring.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (131001341114393) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65500670557046 + 65500670557047.

It is an arithmetic number, because the mean of its divisors is an integer number (65500670557047).

Almost surely, 2131001341114093 is an apocalyptic number.

It is an amenable number.

131001341114093 is a deficient number, since it is larger than the sum of its proper divisors (1).

131001341114093 is an equidigital number, since it uses as much as digits as its factorization.

131001341114093 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 3888, while the sum is 32.

The spelling of 131001341114093 in words is "one hundred thirty-one trillion, one billion, three hundred forty-one million, one hundred fourteen thousand, ninety-three".