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31211452523117 is a prime number
BaseRepresentation
bin1110001100010111110111…
…00010100110001001101101
311002111210012210100022201102
413012023323202212021231
513042332023321214432
6150214202055052445
76400644614053406
oct706137342461155
9132453183308642
1031211452523117
119a437802a5289
123600ba4371125
1314552c5622849
1479c8dc8dc9ad
15391d3513e962
hex1c62fb8a626d

31211452523117 has 2 divisors, whose sum is σ = 31211452523118. Its totient is φ = 31211452523116.

The previous prime is 31211452523099. The next prime is 31211452523161. The reversal of 31211452523117 is 71132525411213.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 16439221629961 + 14772230893156 = 4054531^2 + 3843466^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-31211452523117 is a prime.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (31211452583117) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 15605726261558 + 15605726261559.

It is an arithmetic number, because the mean of its divisors is an integer number (15605726261559).

Almost surely, 231211452523117 is an apocalyptic number.

It is an amenable number.

31211452523117 is a deficient number, since it is larger than the sum of its proper divisors (1).

31211452523117 is an equidigital number, since it uses as much as digits as its factorization.

31211452523117 is an evil number, because the sum of its binary digits is even.

The product of its digits is 50400, while the sum is 38.

The spelling of 31211452523117 in words is "thirty-one trillion, two hundred eleven billion, four hundred fifty-two million, five hundred twenty-three thousand, one hundred seventeen".