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553325213143303 is a prime number
BaseRepresentation
bin111110111001111110001000…
…0110001100101000100000111
32200120011101002110220121012221
41331303330100301211010013
51040011140014031041203
65240454032124553211
7224356266414065056
oct17563742061450407
92616141073817187
10553325213143303
111503409a572a647
12520861723b2807
131a999461188463
149a8d12448a39d
1543e83d2bd6abd
hex1f73f10c65107

553325213143303 has 2 divisors, whose sum is σ = 553325213143304. Its totient is φ = 553325213143302.

The previous prime is 553325213143267. The next prime is 553325213143333. The reversal of 553325213143303 is 303341312523355.

It is a strong prime.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-553325213143303 is a prime.

It is a super-2 number, since 2×5533252131433032 (a number of 30 digits) contains 22 as substring.

It is a self number, because there is not a number n which added to its sum of digits gives 553325213143303.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (553325213143333) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 276662606571651 + 276662606571652.

It is an arithmetic number, because the mean of its divisors is an integer number (276662606571652).

Almost surely, 2553325213143303 is an apocalyptic number.

553325213143303 is a deficient number, since it is larger than the sum of its proper divisors (1).

553325213143303 is an equidigital number, since it uses as much as digits as its factorization.

553325213143303 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 1458000, while the sum is 43.

Adding to 553325213143303 its reverse (303341312523355), we get a palindrome (856666525666658).

The spelling of 553325213143303 in words is "five hundred fifty-three trillion, three hundred twenty-five billion, two hundred thirteen million, one hundred forty-three thousand, three hundred three".