Search a number
-
+
334042118497 is a prime number
BaseRepresentation
bin1001101110001100111…
…01100011100101100001
31011221012222011211021201
410313012131203211201
520433104240242442
6413242401521201
733063606105253
oct4670635434541
91157188154251
10334042118497
1111973708182a
12548a6043201
13256667712a9
141224c345cd3
158a510cc7b7
hex4dc6763961

334042118497 has 2 divisors, whose sum is σ = 334042118498. Its totient is φ = 334042118496.

The previous prime is 334042118491. The next prime is 334042118503. The reversal of 334042118497 is 794811240433.

It is a balanced prime because it is at equal distance from previous prime (334042118491) and next prime (334042118503).

It can be written as a sum of positive squares in only one way, i.e., 327356766801 + 6685351696 = 572151^2 + 81764^2 .

It is a cyclic number.

It is not a de Polignac number, because 334042118497 - 211 = 334042116449 is a prime.

It is a super-3 number, since 3×3340421184973 (a number of 36 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.

It is not a weakly prime, because it can be changed into another prime (334042118491) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 167021059248 + 167021059249.

It is an arithmetic number, because the mean of its divisors is an integer number (167021059249).

Almost surely, 2334042118497 is an apocalyptic number.

It is an amenable number.

334042118497 is a deficient number, since it is larger than the sum of its proper divisors (1).

334042118497 is an equidigital number, since it uses as much as digits as its factorization.

334042118497 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 580608, while the sum is 46.

The spelling of 334042118497 in words is "three hundred thirty-four billion, forty-two million, one hundred eighteen thousand, four hundred ninety-seven".