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130313523497 is a prime number
BaseRepresentation
bin111100101011101001…
…0101001000100101001
3110110100202212220000012
41321113102221010221
54113340230222442
6135510515021305
712262201450013
oct1712722510451
9413322786005
10130313523497
11502a164aa63
122130989a835
13c399b28bb1
146442d9d1b3
1535ca644582
hex1e574a9129

130313523497 has 2 divisors, whose sum is σ = 130313523498. Its totient is φ = 130313523496.

The previous prime is 130313523371. The next prime is 130313523523. The reversal of 130313523497 is 794325313031.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 117723785881 + 12589737616 = 343109^2 + 112204^2 .

It is an emirp because it is prime and its reverse (794325313031) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 130313523497 - 214 = 130313507113 is a prime.

It is a super-2 number, since 2×1303135234972 (a number of 23 digits) contains 22 as substring.

It is not a weakly prime, because it can be changed into another prime (130313523797) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65156761748 + 65156761749.

It is an arithmetic number, because the mean of its divisors is an integer number (65156761749).

Almost surely, 2130313523497 is an apocalyptic number.

It is an amenable number.

130313523497 is a deficient number, since it is larger than the sum of its proper divisors (1).

130313523497 is an equidigital number, since it uses as much as digits as its factorization.

130313523497 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 204120, while the sum is 41.

The spelling of 130313523497 in words is "one hundred thirty billion, three hundred thirteen million, five hundred twenty-three thousand, four hundred ninety-seven".