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53715131231 is a prime number
BaseRepresentation
bin110010000001101010…
…111101101101011111
312010122111111101120212
4302001222331231133
51340002033144411
640402033403035
73611053040066
oct620152755537
9163574441525
1053715131231
1120864878143
12a4b110947b
1350b0660928
142857ccd1dd
1515e5ae968b
hexc81abdb5f

53715131231 has 2 divisors, whose sum is σ = 53715131232. Its totient is φ = 53715131230.

The previous prime is 53715131207. The next prime is 53715131233. The reversal of 53715131231 is 13213151735.

It is a strong prime.

It is a cyclic number.

It is not a de Polignac number, because 53715131231 - 214 = 53715114847 is a prime.

It is a super-2 number, since 2×537151312312 (a number of 22 digits) contains 22 as substring.

Together with 53715131233, it forms a pair of twin primes.

It is a Chen prime.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (53715131233) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 26857565615 + 26857565616.

It is an arithmetic number, because the mean of its divisors is an integer number (26857565616).

Almost surely, 253715131231 is an apocalyptic number.

53715131231 is a deficient number, since it is larger than the sum of its proper divisors (1).

53715131231 is an equidigital number, since it uses as much as digits as its factorization.

53715131231 is an odious number, because the sum of its binary digits is odd.

The product of its digits is 9450, while the sum is 32.

Adding to 53715131231 its reverse (13213151735), we get a palindrome (66928282966).

The spelling of 53715131231 in words is "fifty-three billion, seven hundred fifteen million, one hundred thirty-one thousand, two hundred thirty-one".