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2163737943413 is a prime number
BaseRepresentation
bin11111011111001000110…
…101001101110101110101
321122211222121200012100122
4133133020311031311311
5240422313403142123
64334001341221325
7312216305026334
oct37371065156565
97584877605318
102163737943413
117646aa72a182
122ab41b561845
1312906904b317
1476a229bdd1b
153b43cbc8bc8
hex1f7c8d4dd75

2163737943413 has 2 divisors, whose sum is σ = 2163737943414. Its totient is φ = 2163737943412.

The previous prime is 2163737943337. The next prime is 2163737943431. The reversal of 2163737943413 is 3143497373612.

Together with next prime (2163737943431) it forms an Ormiston pair, because they use the same digits, order apart.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 2033082443044 + 130655500369 = 1425862^2 + 361463^2 .

It is a cyclic number.

It is not a de Polignac number, because 2163737943413 - 240 = 1064226315637 is a prime.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (2163737943443) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1081868971706 + 1081868971707.

It is an arithmetic number, because the mean of its divisors is an integer number (1081868971707).

Almost surely, 22163737943413 is an apocalyptic number.

It is an amenable number.

2163737943413 is a deficient number, since it is larger than the sum of its proper divisors (1).

2163737943413 is an equidigital number, since it uses as much as digits as its factorization.

2163737943413 is an evil number, because the sum of its binary digits is even.

The product of its digits is 6858432, while the sum is 53.

The spelling of 2163737943413 in words is "two trillion, one hundred sixty-three billion, seven hundred thirty-seven million, nine hundred forty-three thousand, four hundred thirteen".