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131435534153593 is a prime number
BaseRepresentation
bin11101111000101000111000…
…010100101111011101111001
3122020101010002100210010011101
4131320220320110233131321
5114211414433220403333
61143312352324531401
736453622030530115
oct3570507024573571
9566333070703141
10131435534153593
113897455156aaaa
12128a9106550b61
135845418c65537
142465729bda745
15102de170eb67d
hex778a3852f779

131435534153593 has 2 divisors, whose sum is σ = 131435534153594. Its totient is φ = 131435534153592.

The previous prime is 131435534153387. The next prime is 131435534153611. The reversal of 131435534153593 is 395351435534131.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 105031588274064 + 26403945879529 = 10248492^2 + 5138477^2 .

It is a cyclic number.

It is not a de Polignac number, because 131435534153593 - 233 = 131426944219001 is a prime.

It is a super-4 number, since 4×1314355341535934 (a number of 58 digits) contains 4444 as substring.

It is not a weakly prime, because it can be changed into another prime (131435534153093) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65717767076796 + 65717767076797.

It is an arithmetic number, because the mean of its divisors is an integer number (65717767076797).

Almost surely, 2131435534153593 is an apocalyptic number.

It is an amenable number.

131435534153593 is a deficient number, since it is larger than the sum of its proper divisors (1).

131435534153593 is an equidigital number, since it uses as much as digits as its factorization.

131435534153593 is an odious number, because the sum of its binary digits is odd.

The product of its digits is 21870000, while the sum is 55.

The spelling of 131435534153593 in words is "one hundred thirty-one trillion, four hundred thirty-five billion, five hundred thirty-four million, one hundred fifty-three thousand, five hundred ninety-three".