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5192631711353 is a prime number
BaseRepresentation
bin100101110010000000011…
…1101111111011001111001
3200101102002102122222002002
41023210000331333121321
51140034002204230403
615013243450350345
71044104153202062
oct113440075773171
920342072588062
105192631711353
111722201258409
126ba449a2b3b5
132b88809bcb18
1413d47903d569
1590113d7e188
hex4b900f7f679

5192631711353 has 2 divisors, whose sum is σ = 5192631711354. Its totient is φ = 5192631711352.

The previous prime is 5192631711341. The next prime is 5192631711413. The reversal of 5192631711353 is 3531171362915.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 4539320002624 + 653311708729 = 2130568^2 + 808277^2 .

It is a cyclic number.

It is not a de Polignac number, because 5192631711353 - 214 = 5192631694969 is a prime.

It is a super-3 number, since 3×51926317113533 (a number of 39 digits) contains 333 as substring.

It is a Sophie Germain prime.

It is a Curzon number.

It is a junction number, because it is equal to n+sod(n) for n = 5192631711298 and 5192631711307.

It is not a weakly prime, because it can be changed into another prime (5192631710353) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2596315855676 + 2596315855677.

It is an arithmetic number, because the mean of its divisors is an integer number (2596315855677).

Almost surely, 25192631711353 is an apocalyptic number.

It is an amenable number.

5192631711353 is a deficient number, since it is larger than the sum of its proper divisors (1).

5192631711353 is an equidigital number, since it uses as much as digits as its factorization.

5192631711353 is an evil number, because the sum of its binary digits is even.

The product of its digits is 510300, while the sum is 47.

The spelling of 5192631711353 in words is "five trillion, one hundred ninety-two billion, six hundred thirty-one million, seven hundred eleven thousand, three hundred fifty-three".