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113143034240377 is a prime number
BaseRepresentation
bin11001101110011100101010…
…010111010101110101111001
3112211121100221120222022121121
4121232130222113111311321
5104312213413231143002
61040345105315305241
732555212601665051
oct3156345227256571
9484540846868547
10113143034240377
1133061767217789
1210833a652ab221
134b1945b1a594b
141dd221b422561
15d1319b232237
hex66e72a5d5d79

113143034240377 has 2 divisors, whose sum is σ = 113143034240378. Its totient is φ = 113143034240376.

The previous prime is 113143034240231. The next prime is 113143034240413. The reversal of 113143034240377 is 773042430341311.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 98834396519401 + 14308637720976 = 9941549^2 + 3782676^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-113143034240377 is a prime.

It is a super-3 number, since 3×1131430342403773 (a number of 43 digits) contains 333 as substring.

It is not a weakly prime, because it can be changed into another prime (113143034250377) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 56571517120188 + 56571517120189.

It is an arithmetic number, because the mean of its divisors is an integer number (56571517120189).

Almost surely, 2113143034240377 is an apocalyptic number.

It is an amenable number.

113143034240377 is a deficient number, since it is larger than the sum of its proper divisors (1).

113143034240377 is an equidigital number, since it uses as much as digits as its factorization.

113143034240377 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 508032, while the sum is 43.

Adding to 113143034240377 its reverse (773042430341311), we get a palindrome (886185464581688).

The spelling of 113143034240377 in words is "one hundred thirteen trillion, one hundred forty-three billion, thirty-four million, two hundred forty thousand, three hundred seventy-seven".